Notes for
Analysis on Manifolds via the Laplacian
by Yaiza Canzani.
Math 253, Fall 2013. Harvard University.
3
Abstract
These are the notes of a course I taught on Fall 2013 at Harvard University. Any comments and suggestions are welcome. I plan on improving these notes next time I teach the course. The notes may have mistakes, so use them at your own risk. Also, many citations are missing. The following references were important sources for these notes:
Eigenvalues in Riemannian geometry. By I. Chavel. • Old and new aspects in Spectral Geometry. By M. Craiveanu, M. Puta and T. Ras- • sias.
The Laplacian on a Riemannian manifold. By S. Rosenberg. • Local and global analysis of eigenfunctions on Riemannian manifolds. By S. Zelditch. • I would like to thank Evans Harrell and Richard Laugesen for sharing with me their thoughts and experiences on teaching courses like this one.
Enjoy!!
Contents
Abstract ...... 3
1 What makes the Laplacian special? 7 1.1 Almost daily life problems ...... 7 1.2 Why not another operator? ...... 9 1.3 You need to solve ∆ϕ = λϕ !...... 9 1.4 A hard problem: understanding the eigenvalues ...... 10 1.4.1 Direct problems ...... 10 1.4.2 Inverse problem: Can you hear the shape of a drum? ...... 12 1.5 An extremely hard problem: understanding the eigenfunctions . . . . . 13
2 Laplacian in Euclidean spaces 19 2.1 Interval ...... 20 2.2 Rectangle ...... 23 2.3 Disc ...... 26 2.4 Harmonic functions ...... 28
3 A very brief review of differential and Riemannian geometry 35 3.1 Differentiable Manifolds ...... 35 3.2 Differential forms ...... 38 3.3 Riemannian Geometry ...... 39
4 The Laplacian on a Riemannian manifold 41 4.1 Definition ...... 41 4.2 Examples ...... 44 4.3 The Laplacian under conformal deformations (Exercise) ...... 45 4.4 The Laplacian commutes with isometries ...... 46 4.5 The Laplacian and Riemannian coverings ...... 46 4.6 The Laplacian on product manifolds ...... 47 4.7 Green’s theorem ...... 48 6 CONTENTS
5 Examples on manifolds 51 5.1 Circle ...... 51 5.2 Torus ...... 51 5.3 Rectangular prisms ...... 53 5.4 Sphere ...... 54 5.5 Klein Bottle (Exercise) ...... 60 5.6 Projective space (Exercise) ...... 60
6 Heat Operator 61 6.1 Sobolev Spaces ...... 61 6.2 Heat propagator ...... 64 6.3 Basis of eigenfunctions on compact manifolds ...... 67 n 6.4 Fundamental solution in R ...... 70 6.5 Existence of the fundamental solution on compact manifolds ...... 72 6.6 Fundamental solution in Riemannian coverings ...... 78
7 Eigenvalues 83 7.1 Self-adjoint extension of the Laplacian ...... 83 7.2 Characterization ...... 86 7.3 Domain monotonicity ...... 88 7.4 Simple lower bound for λk ...... 90 7.5 First eigenvalue: Faber Krahn inequality ...... 91 7.6 Continuity of eigenvalues ...... 93 7.7 Multiplicity of eigenvalues ...... 96 7.8 High energy eigenvalue asymptotics ...... 96 7.9 Isospectral manifolds ...... 99
8 Eigenfunctions 101 8.1 Local properties of Eigenfunctions ...... 102 8.1.1 Gradient estimates ...... 104 8.1.2 Positive mass on subsets of M ...... 105 8.2 Global properties of eigenfunctions ...... 105 8.2.1 L∞-norms ...... 105 8.2.2 Lp-norms...... 107 8.3 Zeros of eigenfunctions ...... 108 8.4 Random wave conjecture ...... 113 CHAPTER 1
What makes the Laplacian special?
In this Chapter we motivate the study of the Laplace operator. To simplify exposition, we do this by concentrating on planar domains.
1.1 Almost daily life problems
n ∞ Let Ω R be a connected domain and consider the operator ∆ acting on C (Ω) that ⊂ simply differentiates a function ϕ C∞(Ω) two times with respect to each position ∈ variable: n X ∂2ϕ ∆ϕ = . − ∂x2 i=1 i This operator is called the Laplacian on Ω. You might also have seen it defined as ∆ = div . This is actually the definition of the Laplacian on a Riemannian manifold − ∇ (M, g). Then the Riemannian Laplacian is defined as
∆g = divg g − ∇ where divg is the divergence operator and g is the gradient one. Here are some exam- ∇ ples where the Laplacian plays a key role:
Steady-state Fluid Flow. Suppose you want to study the velocity v(x1, x2, x3, t) of a given fluid. If the flow is steady, then the velocity field should be independent of the time t. If the flow is irrotational, curl v = 0, then v = u for some function u known −∇ as the velocity potential. If the flow is incompressible, then div v = 0. It then follows that u must satisfy the equation ∆u = 0. A function that satisfies such equation is called a harmonic function. Thus the velocity potential for an incompressible irrotational fluid is harmonic. 8 What makes the Laplacian special?
Static electric field. A static electric field E is governed by the following equations curlE = 0 and divE = 4πρ where ρ denotes the charge density. Since curlE = 0 it fol- lows that E = u for some function known as the electric potential. You then must −∇ have ∆u = 0. In other words, the electric potential in a charge free region is harmonic.
Heat diffusion. If you are interested in understanding how would heat propagate n along Ω R then you should solve the Heat Equation ⊂ 1 ∂ ∆u(x, t) = u(x, t) − c ∂t where c is the conductivity of the material of which Ω is made of, and u(x, t) is the temperature at the point x Ω at time t. ∈ You could also think you have an insulated region Ω (it could be a wire, a ball, etc.) and apply certain given temperatures on the edge ∂Ω. If you want to know what the temperature will be after a long enough period of time (that is, the steady state temperature distribution), then you need to find a solution of the heat equation that be independent of time. The steady state temperature solution will be a function u(x1, . . . , xn, t) such that ∆u = 0. Wave propagation. Now, instead of applying heat to the surface suppose you cover it with a thin layer of some fluid and you wish to describe the motion of the surface of the fluid. Then you will need to solve the Wave equation 1 ∂2 ∆u(x, t) = u(x, t) − c ∂t2 where √c is the speed of sound in your fluid, and u(x, t) denotes the height of the wave above the point x at time t.
You could also think of your domain Ω as the membrane of a drum, in which case its boundary ∂Ω would be attached to the rim of the drum. Suppose you want to study what will happen with the vibration you would generate if you hit it. Then, you 2 should also solve the wave equation ∆u(x, t) = ∂ u(x, t) for your drum, but this time − ∂2t you want to make sure that you take into account that the border of the membrane is fixed. Thus, you should also ask your solution to satisfy u(x, t) = 0 for all points x ∂Ω. ∈ Quantum particles. If you are a bit more eccentric and wish to see how a quantum particle moves inside Ω (under the assumption that there are no external forces) then you need to solve the Schr¨odingerEquation 2 ~ ∂ ∆u(x, t) = i~ u(x, t) 2m ∂t where ~ is Planck’s constant and m is the mass of the free particle. Normalizing u so that u( , t) 2 = 1 one interprets u(x, t) as a probability density. That is, if k · kL (Ω) A Ω then the probability that your quantum particle be inside A at time t is given ⊂ by R u(x, t) 2dx. A | | 1.2 Why not another operator? 9
1.2 Why not another operator?
n The Laplacian on R commutes with translations and rotations. That is, if T is a translation or rotation then ∆(ϕ T ) = (∆ϕ) T . Something more striking occurs, ◦ ◦ if S is any operator that commutes with translations and rotations then there exist Pm j coefficients a1, . . . , am making S = j=1 aj ∆ . Therefore, it is not surprising that the Laplacian will be a main star in any process whose underlying physics are independent n of position and direction such as heat diffusion and wave propagation in R . We will show that on general Riemannian manifolds the Laplacian commutes with isometries.
1.3 You need to solve ∆ϕ = λϕ !
There are of course many more problems involving the Laplacian, but we will focus on these ones to stress the importance of solving the eigenvalue problem (also known as Helmholtz equation) ∆ϕ = λϕ. It is clear that if one wants to study harmonic functions then one needs to solve the equation ∆ϕ = λϕ with λ = 0. So the need for understanding solutions of the Helmholtz equation for problems such as the static electric field or the steady-state fluid flow is straightforward. In order to attack the heat diffusion, wave propagation and Schr¨odingerproblems described above, a standard method (inspired by Stone-Weierstrass Theorem) is to look for solutions u(x, t) of the form u(x, t) = α(t)ϕ(x). For instance if you do this and look at the Heat equation then you must have ∆ϕ(x) α0(t) = x Ω, t > 0. ϕ(x) − α(t) ∈ This shows that there must exist a λ R such that ∈ α0 = λα and ∆ϕ = λϕ. − Therefore ϕ must be an eigenfunction of the Laplacian with eigenvalue λ and α(t) = −λt −λ t e . Once you have these particular solutions uk = e k ϕk you use the superposition principle to write a general solution
X −λkt u(x, t) = ake ϕk(x) k where the coefficients ak are chosen depending on the initial conditions. You could do the same with the wave equation (we do it in detail for a guitar string in Section 2.1) or with the Schr¨odingierequation and you will also find particular solutions of the form uk(x, t) = αk(t)ϕk(x) with e−λkt Heat eqn, √ i λ t ∆ϕk = λkϕk and αk(t) = e k Wave eqn, eiλkt Schr¨odingereqn. 10 What makes the Laplacian special?
1.4 A hard problem: understanding the eigenvalues
Section 1.3 shows why it is so important to understand the eigenvalues λk together with the eigenfunctions ϕk of the Laplacian. The truth is that doing so is a very hard task. Indeed one can only explicitly compute the eigenvalues for very specific choices of regions Ω such as rectangles, discs, ellipses and a few types of triangles (see Section ??). Understanding the eigenvalues is so hard that for hexagons not even the first eigenvalues is known! This brings us to the following question:
Question 1 (Direct problem). If I know (more or less) the shape of a domain, what can I deduce of its Laplace eigenvalues?
On the other side of the road, it wouldn’t be weird to expect the eigenvalues of the Laplacian on Ω to carry some information of the geometry of Ω.
Question 2 (Inverse problem). If I know (more or less) the Laplace eigenvalues of a domain, what can I deduce of its geometry?
1.4.1 Direct problems The first eigenvalue: Rayleigh Conjecture
The first eigenvalue λ1 of the Laplacian on an interval or a region of the plane is called the fundamental tone. This is because either on a vibrating guitar string or drum mem- brane the first eigenvalue corresponds to the leading frequency of oscillation and it is therefore the leading tone you hear when you play one of these instruments. Seen from a heat-diffusion point of view, since the solutions of the heat equation are of the form P −λnt u(x, y, t) = n ane ϕn(x, y), it is clear that (λ1, ϕ1) give the dominant information −λ t because e 1 ϕ1(x, y) is the mode that decays with slowest rate as time passes by. From this last point of view it is natural to expect that the geometry of Ω should be reflected on λ1 to some extent. For instance the largest the boundary ∂Ω is, the more quickly the heat should wear off. That is, if we consider a domain Ω and a ball B of same area as Ω, then we expect the heat on Ω to diffuse more quickly than that of B. Therefore, we should have
Faber-Krahn Inequality: λ1(Ω) λ1(B). ≥ This result was proved by Faber and Krahn in 1923. As expected, it extends to any dimension.
Counting function: Lorentz conjecture Jeans asked once what is the energy corresponding to an infinitesimal frequency interval. In 1966 Mark Kac told this story in a very illustrating manner:
...At the end of October of 1910 the great Dutch physicist H. A. Lorentz was invited to G¨otingento deliver a Wolfskehl lecture... Lorentz gave five lectures under the overall title “Old and new problems of physics” and at 1.4 A hard problem: understanding the eigenvalues 11
the end of the fourth lecture he spoke as follows (in free translation from the original German): In conclusion, there is a mathematical problem which perhaps will arouse the interest of mathematicians who are present. It originates in the radiation theory of Jeans. In an enclosure with a perfectly reflecting surface, there can form stand- ing electromagnetic waves analogous to tones over an organ pipe: we shall confine our attention to very high overtones. Jeans asks for the energy in the frequency interval dν. To this end, he calculates the number of overtones which lie between frequencies ν and ν +dν, and multiplies this number by the energy which belongs to the frequency ν, and which according to a theorem of statistical mechanics, is the same for all frequencies. It is here that there arises the mathematical problem to prove that the number of sufficiently high overtones which lie between ν and ν + dν is independent of the shape of the enclosure, and is simply proportional to its volume. For many shapes for which calculations can be carried out, this theorem has been verified in a Leiden dissertation. There is no doubt that it holds in general even for multiply connected regions. Similar theorems for other vibrating structures, like membranes, air masses, etc., should also hold.
If we express the Lorentz conjecture in a vibrating membrane Ω, it becomes of the following form: Let λ1 λ2 ... be the Laplace eigenvalues corresponding to the ≤ ≤ problem
∆ϕk = λkϕk ϕk ∂Ω = 0. | Then area(Ω) N(λ) = # λk : λk < λ λ as λ . { } ∼ 2π → ∞ D. Hilbert was attending these lectures and predicted as follows: “This theorem would not be proved in my life time.” But, in fact, Hermann Weyl, a graduate student at that time, was also attending these lectures. Weyl proved this conjecture four months later in February of 1911.
We will prove this in specific examples such as rectangles and the torus. Later on we will prove the analogue result for compact Riemannian manifolds (M, g). Let λ0 λ1 ... ≤ ≤ be the Laplace eigenvalues repeated according to its multiplicity. Then
ωn N(λ) V ol(M)λn/2, λ ∼ (2π)n → ∞
n where ωn is the volume of the unit ball in R . In particular,
2 (2π) 2/n λj 2/n j , j . ∼ (ωnV ol(M)) → ∞ 12 What makes the Laplacian special?
1.4.2 Inverse problem: Can you hear the shape of a drum? Suppose you have perfect pitch. Could you derive the shape of a drum from the music you hear from it? More generally, can you determine the structural soundness of an object by listening to its vibrations? This question was first posed by Schuster in 1882. As Berger says in his book A panoramic view of Riemannian Geometry, “Already in the middle ages bell makers knew how to detect invisible cracks by sounding a bell on the ground before lifting it up to the belfry. How can one test the resistance to vibrations of large modern structures by non- destructive essays?... A small crack will not only change the boundary shape of our domain, one side of the crack will strike the other during vibrations invalidating our use of the simple linear wave equation. On the other hand, heat will presumably not leak out of a thin crack very quickly, so perhaps the heat equation will still provide a reasonable approximation for a short time...” Infinite sequences of numbers determine via Fourier analysis an integrable function. It wouldn’t be that crazy if an infinite sequences of eigenvalues would determine the shape of the domain. Unfortunately, the answer to the question can you hear the shape of a drum? is no. This was proved in 1992 by Gordon, Web and Wolpert [?]. Nowadays many planar domains are known to have different shapes but exactly the same spectrum.
Figure: Two domains with the same eigenvalues Picture from the paper LaplaceBeltrami spectra as Shape-DNA of surfaces and solids Not all is lost. One can still derive a lot of information of a domain by knowing its eigenvalues. Using the heat kernel, in 1966 Mark Kac proved the formula ∞ X 1 2πt e−λkt area(Ω) √4πt length(∂Ω) + (1 γ(Ω)) ∼ 4πt − 3 − k=1 where γ(Ω) is the genus of Ω and Ω is a polygon. The eigenvalues λn are the ones corre- sponding to the Laplacian on Ω enforcing ϕk ∂Ω = 0. A year later McKean and Singer | (1967) proved the same result in the context of Riemannian manifolds with boundary.
This means that if you know the full sequence of eigenvalues of your favorite domain Ω then you can deduce its area, its perimeter and the number of holes in it!!
On a compact Riemmanian manifold without boundary Minakshisundaram (1953) proved the analog weaker result ∞ Z X −λkt 1 t 2 e n vol(M) + Rg(x) ωg + O(t ) , ∼ 2 6 k=1 (4πt) M 1.5 An extremely hard problem: understanding the eigenfunctions 13
where Rg denotes the scalar curvature. So you can hear the dimension, the volume and the total scalar curvature of a compact Riemannian manifold.
1.5 An extremely hard problem: understanding the eigen- functions
Eigenfunctions of the Laplacian play a key role whenever it comes to do analysis on Riemannian manifolds. One of the main reasons is that they are the key ingredient to carry an analog of Fourier series on manifolds. Indeed, as we shall prove later, we have
Sturm-Liouville’s decomposition. Give a compact Riemannian manifold (M, g) there is an orthonormal basis ϕ1, . . . , ϕj,... of eigenfunctions of the Laplacian ∆g with 2 respective eigenvalues 0 λ1 λj ... such that any function φ L (M) can ≤ ≤ · · · ≤ ≤ ∈ be written as a convergent series in L2(M)
∞ X φ = ajϕj j=1
for some coefficients aj R. ∈
On the other hand, as already mentioned, we may interpret the eigenfunction ϕj as the probability density of a quantum particle in the energy state λj. That is, the probability that a quantum particle in the state ϕj belongs to the set A M if given by ⊂
Z 2 ϕj ωg. A | |
High energy eigenfunctions are expected to reflect the dynamics of the geodesic flow. In the energy limit λ one should be able to recover classical mechanics from → ∞ quantum mechanics. In the following picture (taken from Many-body quantum chaos: Recent developments and applications to nuclei) you can see how the dynamics of the geodesic flow for two different systems is reflected on the eigenfunctions. In the left column a cardioid billiard is represented. In the right column a ring billiard is shown. In the first line the trajectories of the geodesic flow for each system is shown. Then, 2 from the second line to the fifth one, the graph of the functions ϕj is shown for | | λj = 100, 1000, 1500, 2000. The darker the color, the higher the value of the modulus. One can see how a very chaotic system, such as the cardioid, yields a uniform distribution (chaotic) of the eigenfunctions. On the other hand, a very geometric dynamical system, such as the ring, yields geometric distributions of the eigenfunctions. 14 What makes the Laplacian special? 64 reflect the structure of the corresponding classical dynamics. integrable Figure circular 4. billiard Behavior and of the eigenstates. chaotic The cardioid eigenstates billiard of the displayed in the last column for each case. clearly seen in the quantum Poincaré Husimi representation two lines) or extend over the chaotic region (last line). This is most either concentrate in the regular islands (first phase space typically Figure 5. Mixed phase space. Eigenstates in billiards with mixed
–1 p –1 p –1 p 0 1 0 1 0 1 404 0 –4 4 0 –4 4 0 –4 s s s sität Ulm. Contact him at [email protected]. Bäcker has a PhD in theoretical physics from the Univer- ear dynamics, quantum chaos, and mesoscopic systems. Universität Dresden. His research interests include nonlin- Arnd Bäcker References 0 A. Bäcker, R. Ketzmerick, and A. Monastra, “Flooding of Regular 10. .R. Aurich and F. Steiner, “Statistical Properties of Highly Excited 4. Eigenfunction and Eigenvalue of Aspects “Numerical Bäcker, A. 3. A. Bäcker and H.R. Dullin, “Symbolic Dynamics and Periodic2. Orbits M. Robnik, “Classical Dynamics of a Family of Billiards with Ana- 1. .M.L. Mehta, 7. M.V. Berry and M. Tabor, “Level Clustering in the Regular Spec- 6. .A. Bäcker, R. Schubert, and P. Stifter, “Rate of Quantum Ergod- 9. Irregular and Regular of Mechanics “Semiclassical Berry, M.V. 8. O. Bohigas, M.-J. Giannoni, and C. Schmit, “Characterization of 5. Quantum Eigenstates of a Strongly Chaotic System,” Esposti and S. Graffi, eds., Springer-Verlag, 2003, pp. 91–144. Aspects of Quantum Maps Computations for Chaotic Quantum Systems,” for the Cardioid Billiard,” lytic Boundaries,” Islands by Chaotic States,” 5425–5447. Billiards,” Euclidean in icity mann, and R. Stora, eds., North-Holland, 1983, pp. 171–271. Chaotic Behaviour of Deterministic Systems Motion,” trum,” Laws,” Chaotic Quantum Spectra and Universality of Level Fluctuation vol. 64, 1993, pp. 185–214. 054102. Physical Rev. Letters Proc. Royal Soc. London A Comportement Chaotique des Systemes Deterministes: is a scientific researcher at the Technische Random Matrices J. Physics A C MUIGIN OMPUTING J. Physics A , Lecture Notes in Physics 618, M. Degli , vol. 52, 1984, pp. 1–4. Physical Rev. Letters , vol. 16, 1983, pp. 3971–3986. Physical Rev. E Rev. Physical , Academic Press, 1991. , vol. 356, 1977, pp. 375–394. , vol. 30, 1997, pp. 1991–2020. S CIENCE , G. Iooss, R.H.G. Helle- , vol. 57, 1998, pp. 1998, 57, vol. , , vol. 94, 2005, p. The Mathematical & E NGINEERING Physica D ,
A beautiful result about the behavior of eigenfunctions takes place on manifolds with ergodic geodesic flow (like the cardioid above), including all manifolds with negative constant sectional curvature. This result says that in the high energy limit eigenfunc- tions are equidistributed.
Quantum ergodicity. If (M, g) is a compact manifold with ergodic geodesic flow then there exists a density one subsequence of eigenfunctions ϕj k such that for any { k } A M ⊂ Z 2 vol(A) lim ϕjk ωg = . k→∞ M vol (M) 1.5 An extremely hard problem: understanding the eigenfunctions 15
#{k: jk≤m} By density one subsequence it is meant that infm m = 1. This result is due to Schnirelman (1973) finished by Colin de Verdiere (1975).
3 Image processing. One may describe a give surface M R by a function such as ⊂ the normal vector. That is, to every point x M you associate the normal vector at x ∈ x (n1(x), n2(x), n3(x)). 7→ P∞ (j) Each function nj : M R can be rewritten as an infinite series nj = a ϕi by the → i=1 i Sturm-Liouville decomposition Theorem. You may then truncate the series and work with the approximates manifold described by the function
N N N ! X (1) X (2) X (3) x a ϕi(x), a ϕi(x), a ϕi(x) . 790 G. Rong et al. 7→ i i i i=1 i=1 i=1
Fig. 2a–f. The inversea. M =Dragon. manifold harmonicsb-f. reconstructed transform manifold for models using ofN different= 100, numbers200, 300, 500 of bases. and 900a is eigenfunctions the original dragon model, and b–f are the results of the inverse manifold harmonics transform using m 100, 200, 300, 500 and 900 bases, respectively respectively. Picture from paper Spectral= mesh deformation.
B ( t Thist )/ is2 a way of encoding the geometry of a surfaceresult is to a some vector extentx1, x2 using,...xm as,witheachitem little infor- xk corres- ij ! [ ˜ ˜ ˜ ] k ˜ = | |+|mation| as you want (at the risk of having a worseponding approximation). to each frequency In practicebasis H : the the Bii t /6. = way| people| have of computing eigenfunctions on the dragon isn to discretize it and work t St(i) k k % ∈&with a' discretized version of the Laplacian andxk itsx corresponding, H xi D eigenfunctions.ii H . For (7) ˜ = $ % = i t, t are theinstance, two triangles the method that ofshare approximating the edge (i a, j surface), by a finite numberi 1 of eigenfunctions is ! &= t and t areused their to areas, performβ a, β changeare theof the two position angles op- of some part of the surface. Suppose you have ! i, j i!, j The inverse MHT maps the descriptor from frequency do- | | | | an armadillo standing on two legs (figure a) and you wish to lift one of its legs (figure posite to the edge (i, j),andSt(i) is the set of triangles main onto space domain by reconstructing x at vertex i incident to vertexe) reducingi. as much as possible the amount ofusing computations the first m thatfrequencies: need to be carried to The aboveget formulation the final result. can be What further people simplified are doing into: is to compute the first 99 eigenfunctions on the (discretized) armadillo (figure b) and approximatem the armadillo by them (figure D 1 Qhk λ hk, (6) k − k xi xk H . (8) − =c). Then, you apply the transformation to the approximate= ˜ i armadillo (figure d). Do- k 1 where D is aing diagonal this is matrix much defined cheaper as -computation follows: wise- than&= applying the transformation to the Figure 2 shows the inverse MHT for the Dragon model Dii Bij t /3. = = | | using 100, 200, 300, 500, and 900 bases, respectively. j t St(i) & % ∈& ' The solution to this eigenproblem yields a series of k eigenpairs (H , λk) called the manifold harmonics bases 4 Deformation process (MHB). These bases are orthogonal, i.e. the functional in- ner product Hi, H j 0ifi j.Wealsoensurethatthe The manifold harmonics transform can help us to trans- MHB is orthonormal,$ % = by dividing&= each basis vector Hk by fer the geometric representation of a surface model and its functional norm Hk, Hk .ByusingtheArnoldimethod, its deformation functions from the space domain to the it is possible to compute$ eigenvectors% band-by-band utiliz- frequency domain. When reconstructing geometric and ing the shift-invert spectral transform. A detailed derivation deformation information using only the first m frequen- of these formulae can be found in [21]. cies, we are in fact conducting a low-pass signal filter- Using the MHB, we can define the manifold harmonics ing. The high frequencies (details) are filtered, and we transform (MHT) to convert the geometry of the mani- get a smoother approximation of the original model. In fold surface M into the spectral domain. The geometry x this paper we use a linear Laplacian-based variational al- (resp. y, z)ofthetriangulatedsurfacecanbeconsidered gorithm for mesh deformation. It is well known that the as a piecewise linear function defined over the hat func- gradient-based transformation cannot handle the local ro- n tions φi x i 1 xiφi where xi denotes the x coordinate tation of details [5]. We can overcome this problem by at vertex: i.TheMHTistheprojectionof= = x onto the or- using a spectral multi-resolution approach. The multi- thonormal MHB( via the functional inner product, and the resolution here means different levels of resolution in 16 What makes the Laplacian special?
original armadillo. You may then add all the details to the transformed armadillo by 788 G. Rong et al. an algorithm called.
788 G. Rong et al.
Fig. 1a–e. The spectral mesh deformation pipeline. a The original Armadillo model, b the manifold harmonics bases, c the smoothed model reconstructed using the first m bases, d the deformed smoothed model, and e the deformed model with details added back
added back easily [6]. However, the size of the linear and spectral methods. Section 3 introduces the concept of system is not reduced since the number of unknowns manifold harmonics. The details of our new algorithm are (vertex positions) remains the same for the smoothed sur- given in Sect. 4, and the experimental results are shown in faces. Another approach is to use topological hierarchies Sect. 5. Section 6 concludes the paper with some possible of coarser and coarser meshes [8]. Subdivision surfaces directions of future work. provide a nice coupling of the geometric and topological hierarchy, and the number of unknowns in the coarse sub- division mesh could be significantly smaller than the ori- 2 Related work ginal mesh [2, 27]. However, automatically building sub- division surfaces for arbitrary irregular meshes is a highly Mesh deformation is an active research area in computer Fig. 1a–e. The spectral mesh deformation pipeline. a The original Armadillonon-trivial model,task, sincePictureb the the from originalmanifold the article irregularSpectral harmonics mesh mesh deformation bases, may. notc thegraphics. smoothed There are numerous previous papers on this topic. model reconstructed using the first m bases, d the deformed smoothedhaveAnother model, coherent way and of subdivisione understandingthe deformed connectivity. the behavior model of with eigenfunctions details is added to study backEnergy their nodal minimization has long been a general approach to sets.In this The setpaper, of points we propose where an eigenfunction to use the vanishesspectrum is like of the skeletondeform of the smooth surfaces [3, 22]. A variational approach is Laplace–Beltramieigenfunction. Let us operator write defined on manifold surfaces, introduced in [2] to deform subdivision surfaces. To pre- i.e. manifold harmonics, to compactly encode the defor- serve surface details, they optimize the energy of a deforma- ϕ := x M : ϕj(x) = 0 . mation functions. The manifoldN j { ∈ harmonics can} be pre- tion vector field instead of the deformation energy of vertex added back easily [6]. However, the size of the linear and spectral methods. Section 3 introduces the concept of computedNodal sets playon arbitrary the role of the irregular skeleton meshes. of your manifold. Compared They encode with severalpositions. aspects Multiresolution mesh editing techniques [11, 28] system is not reduced since the number of unknowns othermanifoldof the subspace geometry harmonics. of deformation the manifold. The techniquesThis details is illustrated [9, of 27],in our the the following new com- algorithm figurehave where been the are developed for detail-preserving deformations by (vertex positions) remains the same for the smoothed sur- putationgivennodal sets in of of Sect. somemanifold eigenfunctions 4, and harmonics the on the experimental is armadillo fully are automatic, colored results in blue. and aredecomposing shown in a mesh into several frequency bands. A de- faces. Another approach is to use topological hierarchies theseSect. orthogonal 5. Section bases 6 provide concludes a compact the paper parametrization with someformed possible mesh is obtained by first manipulating the low- for the space of functions defined on the surfaces. We frequency mesh and later adding back the high frequency of coarser and coarser meshes [8]. Subdivision surfaces candirections use very smallof future number work. (compared to the number of details as displacement vectors. provide a nice coupling of the geometric and topological vertices) of frequency components to represent the geom- Yu et al. [23] apply the widely used Poisson equa- hierarchy, and the number of unknowns in the coarse sub- etry and motion of the smoothed model. So the number tion on the 3D model deformation. They set the gradients division mesh could be significantly smaller than the ori- of unknowns in the linear system for the deformation can before and after the deformation to be equal to get a Pois- be2 greatly Related decreased, work to allow interactive manipulation on son equation, which is a linear system. The solutions of ginal mesh [2, 27]. However, automatically building sub- large triangle meshes. The process of our algorithm is de- this equation give the deformed model. This class of al- division surfaces for arbitrary irregular meshes is a highly scribedMesh as deformation the following steps: is an active research area ingorithm computer is called gradient domain deformation.Gradient domain mesh deformation techniques [1, 9, 12, 17, 23, 24, non-trivial task, since the original irregular mesh may not 1.graphics. compute There manifold are harmonics numerous bases previous for the papers original on this topic. 26, 27] have been intensively investigated for mesh edit- have coherent subdivision connectivity. Energymodel with minimizationn vertices (Fig. has 1b); long been a general approach to ing. The main challenge is to handle nontrivial transform- 2. perform the inverse manifold harmonics transform In this paper, we propose to use the spectrum of the deform smooth surfaces [3, 22]. A variational approachations which is include rotations (especially large rotations) using the first m (m n)frequenciestogetasmoothed Laplace–Beltrami operator defined on manifold surfaces, introduced in [2] to deform subdivision surfaces.while To preserving pre- as much as possible the visual character- model (Fig. 1c); ! istic of the shape at interactive rates. The most important i.e. manifold harmonics, to compactly encode the defor- 3.serve perform surface the deformation details, they on optimize the smoothed the energymodel by of a deforma- idea is to factor out the rotation from the deformation. For mation functions. The manifold harmonics can be pre- tionsolving vector a linear field system instead with ofm theunknowns deformation (Fig. 1d); energy of vertex shape editing, the factorization and shape definition have 4. add the details back to the deformed smoothed model computed on arbitrary irregular meshes. Compared with positions. Multiresolution mesh editing techniquesto be[11, solved 28] simultaneously [17] since the target shape is to get the final result (Fig. 1e). other subspace deformation techniques [9, 27], the com- have been developed for detail-preserving deformationsnot explicitly by given. Instead of factoring out the rotation, putation of manifold harmonics is fully automatic, and decomposingThe rest of this a paper mesh is into organized several as follows: frequency Sect. 2bands.abettersolutionistorepresenttheshapewithrotation- A de- these orthogonal bases provide a compact parametrization reviewsformed some mesh previous is obtained work related by to first mesh manipulating deformation invariant the low- coordinates [12]. Zayer et al. [24] use a harmonic for the space of functions defined on the surfaces. We frequency mesh and later adding back the high frequency can use very small number (compared to the number of details as displacement vectors. vertices) of frequency components to represent the geom- Yu et al. [23] apply the widely used Poisson equa- etry and motion of the smoothed model. So the number tion on the 3D model deformation. They set the gradients of unknowns in the linear system for the deformation can before and after the deformation to be equal to get a Pois- be greatly decreased, to allow interactive manipulation on son equation, which is a linear system. The solutions of large triangle meshes. The process of our algorithm is de- this equation give the deformed model. This class of al- scribed as the following steps: gorithm is called gradient domain deformation.Gradient domain mesh deformation techniques [1, 9, 12, 17, 23, 24, 1. compute manifold harmonics bases for the original 26, 27] have been intensively investigated for mesh edit- model with n vertices (Fig. 1b); ing. The main challenge is to handle nontrivial transform- 2. perform the inverse manifold harmonics transform ations which include rotations (especially large rotations) using the first m (m n)frequenciestogetasmoothed while preserving as much as possible the visual character- model (Fig. 1c); ! istic of the shape at interactive rates. The most important 3. perform the deformation on the smoothed model by idea is to factor out the rotation from the deformation. For solving a linear system with m unknowns (Fig. 1d); shape editing, the factorization and shape definition have 4. add the details back to the deformed smoothed model to be solved simultaneously [17] since the target shape is to get the final result (Fig. 1e). not explicitly given. Instead of factoring out the rotation, The rest of this paper is organized as follows: Sect. 2 abettersolutionistorepresenttheshapewithrotation- reviews some previous work related to mesh deformation invariant coordinates [12]. Zayer et al. [24] use a harmonic 1.5 An extremely hard problem: understanding the eigenfunctions 17
Picture from the article Laplace-Beltrami eigenfunctions towards an algorithm that “understands” the Figure 3. Contours of the first eigenfunctions.geometry Note how. the protrusions and symmetries are cap- tured. The eigenfunctions combine the geometric and topological information contained in the shape signal. From a quantum mechanics point of view, nodal sets can be interpreted as the least R 2 likely place for a quantum particle in the state ϕj to be. This is because ϕj ωg = 0. Nϕj basis. In our specific problem, we want to find an approxi- mation of the eigenfunctions of Laplace-Beltrami (Equation One of the main problems about nodal sets is estimating2). The projected their equation size. is givenA famous by: conjecture on this matter is known as Yau’s conjecture on nodal set’s sizes. It says that there i, < ∆f,Φ >= λ
CHAPTER 2
Laplacian in Euclidean spaces
Spectrum of the Laplacian
The aim of this chapter is to compute explicitly in some special easy cases the eigenvalues and eigenfunctions of the Laplacian operator subject to different boundary conditions. We first do this in one dimensional spaces (segments and circles). We then solve the wave equation of an interval in detail and explain the importance of the first eigenvalue and how it relates to the length of the interval. We then study two dimensional domains such as rectangles and discs. We then prove Weyl asymptotics for the rectangle and show how they encode the area of it.
Given a domain Ω with boundary (the reader might think of it as an interval, or a membrane, or an arbitrary manifold) there are two important boundary conditions one may impose on the solutions ϕ of the eigenvalue problem ∆ϕ = λϕ :
Dirichlet boundary conditions: ϕ ∂Ω = 0. |
2 This is used for instance when your domain Ω R is a membrane and you fix its ⊂ boundary as if Ω was a drum. Since you don’t have any vibrations on the rim of a drum you must have ϕ ∂Ω = 0. |
Neuman boundary conditions: ∂νϕ ∂Ω = 0. | Here ν is the unit outward normal vector to the boundary ∂Ω. This condition is used for example when a surface has a prescribed heat flux, such as a perfect insulator (the heat doesn’t go through the boundary, when it hits it it stays in). 20 Laplacian in Euclidean spaces
2.1 Interval
Consider an interval [0, `]. Dirichlet boundary conditions: ϕ(0) = ϕ(`) = 0. −
The eigenfunctions are kπ ϕk(x) = sin x for k 1 ` ≥ kπ 2 with eigenvalues λk = for k 1. ` ≥
ϕ1
ϕ2
ϕ3
ϕ4
Neumann boundary conditions: ϕ0(0) = ϕ0(`) = 0. − The eigenfunctions are kπ ϕk(x) = cos x for k 1 ` ≥ kπ 2 with eigenvalues λk = for k 0. ` ≥ Observations. 1 Note that if we scale our domain by a factor a > 0 we get λk (0, a`) = a2 λk (0, `) . 2 Intuitively, the eigenvalue λ must balance d , and so λ (length scale)−2. We also dx2 ∼ note that we have the asymptotics 2 λk C k ∼ where C is a constant independent of k. For λ > 0 consider the eigenvalue counting function N(λ) = # eigenvalues λ . { ≤ } Proposition 1. (Weyl’s law for intervals) Write λj for the Dirichlet or Neumann eigenvalues of the Laplacian on the interval Ω = [0, `]. Then, length(Ω) N(λ) √λ. ∼ π 2.1 Interval 21
Proof. k2π2 ` N(λ) = max k : λk < λ = max k : < λ √λ. { } `2 ∼ π
Hear the length of a guitar string (solving the wave equation) After the first half of the 18th century mathematicians such as d’Alembert and Bernoulli developed the theory of a vibrational string. As one should expect, the vibrations of a string will depend on many factors such us its length, mass and tension. To simplify our exposition consider a guitar string of length ` which we model as the interval [0, `]. Assume further that the density mass and the tension are constant and equal to 1. Today it comes as no surprise that the behavior of a vibrating string is described by the wave equation. That is, if we write x for a point in the string [0, `] and t for the time variable, then the height u(x, t) of the string above the point x after a time t should satisfy the wave equation ∂2 ∆u(x, t) = u(x, t). −∂t2 There are infinitely many solutions to this problem. But we already know that there are constraints to this problem that we should take into account since the endpoints of the string are fixed and so u(x, t) must satisfy u(0, t) = 0 = u(`, t) for all time t. In addition having a unique solution to our problem depends upon specifying the initial shape of the string f(x) = u(x, 0) and its initial velocity g(x) = ∂tu(x, 0). All in all, we are solving the system
∂2 ∂2 2 u(x, t) = 2 u(x, t) x [0, `], t > 0, − ∂x − ∂t ∈ u(0, t) = 0 = u(`, t) t > 0, u(x, 0) = f(x) x [0, `], ∈ ∂tu(x, 0) = g(x) x [0, `]. ∈ A general sulution of this problem has the form
∞ X u(x, t) = αk(t)ϕk(x), (2.1) k=1 where kπ kπ α (t) = a cos t + b sin t . k k ` k `
The coefficients are ak = f, ϕk and bk = g, ϕk . h i h i
The functions ϕk are called harmonic modes for the string [0, `]. Since fk = kπ is the kπ frequency of the wave ϕk(x) = sin ` x the connection between the eigenvalues λk and the frequencies fk of the harmonic modes of the string is obvious:
1 p f = λ . k 2π k 22 Laplacian in Euclidean spaces
Therefore, the higher the eigenvalue, the higher the frequency is.
Consider the Fourier transform of a function ϕ as Z ∞ ϕ(ξ) = ϕ(x)e−2πixξ. −∞ Digital Analysis of Geophysical Signkπals and Waves The Fourier transform of the function ϕk(x) = sin ` x is 3.4. Examples of the Fourier Transform rπ kπ kπ At this point let us pause( ϕtok m)(akξe) su =re iwe undeδrstaξnd what is proδducξed+ when the f.requency domain representation of s(t) iFs obtained by Fourier 2transform− (equ`ation− 3.1.2 or 3.1.4`). Figure 3.5 contains a simple composite signal in the time domain made up of three separate cosinusoids. Assuming that kπ In tht followinghese cosinuso pictureids extendthe in tim graphse from - i ofnfinsikty( txo )+ =infiAnitky,cos the Fou`riexr trafornsfokrm= of 1th,i2s c,o3m andposite in the last signal yields the result depicted in Figure 3.6. line we put the graph of s1 + s2 + s3.
Picture from www-rohan.sdsu.edu/ jiracek/DAGSAW/3.4.html. Figure 3.5. Three cosine waves with amplitudes A1, A2, and A3 combine to In the following pictureform the a co Fouriermposite s transformignal with am ofpli thetude functionA1 + A2 + As31. + s2 + s3 is shown.
Picture from www-rohan.sdsu.edu/ jiracek/DAGSAW/3.4.html. Figure 3.6. Fourier transform of three-cosine composite signal in Figure 3.5 yields three pairs of real, even delta functions with corresponding amplitudes A1/2, A2/2, and A3/2.
Since the original signal is real and even (cosine functions are clearly even functions), the Fourier transform must be real and even. Three pure cosine oscillations summate to make up s(t) so only three spectral lines are present in the Fourier transform, S(f). These spikes can be represented by Dirac delta functions that are functions of frequency, not of time as we defined in Section 2.3. For example, the Fourier transform of A1cos 2πf1t is
(3.4.1)
This reveals an interesting aspect of the Fourier transform that we avoided talking about earlier, namely that there are values (spectra lines) at both positive and negative frequencies. In this case they appear where the delta functions are non-zero, i.e., where their arguments are zero, at f = +f1 and f = -f1.
The concept of negative frequencies is not widely understood, even though the proper handling of this concept is critical for practical applications of digital processing in the frequency domain. Therefore, we are compelled to convince you of the validity of both positive and negative frequencies so you will appreciate the subtleties when working with them. This we will do in Appendix B. First let’s see what the Fourier transforms are of several of the functions that we’ve encountered so far. The Fourier 2.2 Rectangle 23
P∞ If you pluck a guitar string then you obtain a wave of the form u(x, t) = k=1 αk(t)ϕk(x), and by applying the Fourier transform to it you get
r ∞ π X kπ kπ (u( , t))(ξ) = i αk(t) δ ξ δ ξ + . F · 2 − ` − ` k=1
So you recover all the relevant frequencies and hence all the eigenvalues.
2.2 Rectangle
Consider a rectangle Ω = [0, `] [0, m]. Separate variables using carthesian coordinates × x and y. That is, look for solutions of the form ϕ(x, y) = f(x)g(y).
Dirichlet boundary conditions: ϕ ∂Ω = 0. − |
The eigenfunctions are
jπ kπ ϕ (x, y) = sin x sin y for j, k 1, jk ` m ≥ and have eigenvalues
jπ 2 kπ 2 λjk = + forj, k 1. ` m ≥
Neumann boundary conditions: ∂νϕ ∂Ω = 0. − |
The eigenfunctions are
jπ kπ ϕ (x, y) = cos x cos y for j, k 0, jk ` m ≥ and have eigenvalues
jπ 2 kπ 2 λjk = + forj, k 0. ` m ≥
In the following picture the eigenfunctions on a square are shown. Laplace eigenfunctions on a rectangle
24 Laplacian in Euclidean spaces
LaplaceIn the following eigenfunctions picture the eigenfunctions on a rectangle on a square are shown as if the were seen Mathfrom 319 Lecture above. 27 Mar 14 4 / 9
Math 319 Lecture 27 Mar 14 4 / 9
Observation 1 Note that if we scale our domain by a factor a > 0 we get λk aΩ) = a2 λk Ω and so the eigenvalue λ must balance ∆. Again, λ (length scale)−2. ∼ 2.2 Rectangle 25
Hearing the area of a rectangle (computing Weyl asymptotics) For λ > 0 consider the eigenvalue counting function
N(λ) = # eigenvalues λ . { ≤ }
Proposition 2. (Weyl’s law for rectangles) Write λj for the Dirichlet eigenvalues of the Laplacian on the rectangle Ω = [0, `] [0, m]. Then, × area(Ω) N(λ) λ. ∼ 4π Proof.
n jπ 2 kπ 2 o N(λ) = # (j, k) N N : + λ = # (j, k) N N :(j, k) Eλ ∈ × ` m ≤ ∈ × ∈ 2 2 where E is the first quadrant of the ellipse √ x + √ y 1. To each point λ λ`/π λm/π ≤ (j, k) E with the square ∈ λ
Rj,k = [j 1, j] [k 1, k]. − × − √λm π
Eλ
√λ` π
Since all these squares lie inside Eλ we get 1 area(Ω) N(λ) area(E ) = π(√λ`/π)(√λm/π) = λ. ≤ λ 4 4π Also, the union of the squares covers a copy, E˜ , of E translated by ( 1, 1): λ λ − − [ E˜ (x, y): x 0, y 0 Rj,k. λ ∩ { ≥ ≥ } ⊂ (j,k)∈Eλ
√λm π
E˜λ
√λ` π 26 Laplacian in Euclidean spaces
Comparing areas shows that 1 N(λ) π(√λ`/π)(√λm/π) √λ`/π √λm/π ≥ 4 − − `m ` + m = λ √λ 4π − π area(Ω) perimeter(Ω) = λ √λ. 4π − 2π
area(Ω) Since N(λj) = j, Proposition 2 yields j λj and so we obtain ∼ 4π Corollary 3. Write λ1 λ2 ... for the Dirichlet eigenvalues of the Laplacian on ≤ ≤ the rectangle Ω = [0, `] [0, m]. Then, × 4πj λj as j . ∼ area(Ω) → ∞
2.3 Disc
2 Consider a disc Ω = x R : x 1 . { ∈ | | ≤ } In order to compute the eigenfunctions on the disc we need to separate variables us- ing polar coordinates r and θ. We first derive a formula for the Laplacian in polar p coordinates. Since r = x2 + y2 and tan θ = y/x, one has
∂ϕ ∂ϕ x ∂ϕ y = , ∂x ∂r r − ∂θ r2 ∂ϕ ∂ϕ y ∂ϕ x = + . ∂y ∂r r ∂θ r2
Therefore,
∂2ϕ x2 ∂2ϕ xy ∂2ϕ y2 ∂2ϕ y2 ∂ϕ xy ∂ϕ = 2 + + + 2 ∂x2 r2 ∂r2 − r3 ∂r∂θ r4 ∂θ2 r3 ∂r r4 ∂θ and
∂2ϕ y2 ∂2ϕ xy ∂2ϕ x2 ∂2ϕ x2 ∂ϕ xy ∂ϕ = + 2 + + 2 . ∂y2 r2 ∂r2 r3 ∂r∂θ r4 ∂θ2 r3 ∂r − r4 ∂θ It follows that the Laplacian applied to ϕ has the form
∂2ϕ ∂2ϕ ∂2ϕ 1 ∂ϕ 1 ∂2ϕ ∆ϕ = = . − ∂x2 − ∂y2 − ∂r2 − r ∂r − r2 ∂θ2 Therefore, the Laplacian in polar coordinates takes the form
∂2 1 ∂ 1 ∂2 ∆ = + + . − ∂r2 r ∂r r2 ∂θ2 2.3 Disc 27
We now look for solutions of the form ϕ(r, θ) = R(r)Φ(θ). From ∆ϕ = λϕ we get
1 1 R00(r)Φ(θ) + R0(r)Φ(θ) + R(r)Φ00(θ) = λR(r)Φ(θ) − r r2 and therefore
r2 1 1 Φ00(θ) R00(r)Φ(θ) + R0(r)Φ(θ) + R(r)Φ00(θ) = . R(r) r r2 − Φ(θ)
This means that there exists k such that
Φ00(θ) = k2Φ(θ) − and 1 k2 R00(r)Φ(θ) + R0(r)Φ(θ) + λ R(r) = 0. r − r2
Set x = √λr and J(x) = R(x/√λ). Then,
x2J 00(x) + xJ 0(x) + (x2 k2)J(x) = 0 − which is known as Bessel’s equation. The solution for it is the k-th Bessel function
∞ X ( 1)` xk+2` J (x) = − . k `!(k + `)! 2 `=0
Since R(r) = Jk(√λ r), we get that
λ ϕk(r, θ) = Φk(θ)Jk(√λ r)
λ are eigenfunctions of ∆ where Φk(θ) = ak cos(kθ) + bk sin(kθ). The eigenvalue of ϕk has eigenvalue λ.
Let us now impose boundary conditions:
Dirichlet boundary conditions: We ask ϕλ(1, θ) = 0 for all θ [0, 2π]. This implies k ∈ Jk(√λ) = 0 and so √λ must be a zero of the k-Bessel function.
λ Neumann boundary conditions: We ask ∂rϕk(1, θ) = 0 for all θ [0, 2π]. This implies 0 ∈ Jk(√λ) = 0 and so √λ must be a zero of the derivative of the k-Bessel function.
In the figure below, the first eigenfunctions on the disk are shown. Laplace eigenfunctions on the disk
28 Laplacian in Euclidean spaces
Laplace eigenfunctions on the disk
MathThe 319 Lecture following 30 are the eigenfunctions shown as if they where seen from above.Mar 21 7 / 7
Math 319 Lecture 30 Mar 21 7 / 7 2.4 Harmonic functions
n 2 Let Ω R be an open connected region. A real valued function ϕ C (Ω) is said to ⊂ ∈ be harmonic if ∆ϕ = 0. 2.4 Harmonic functions 29
The theory of harmonic functions is the same as the theory of conservative vector fields n with zero divergence. Indeed, for any vector field F in a connected region Ω R one ⊂ has that one has curlF = 0 and divF = 0 if and only if there exists a harmonic potential ϕ making F = ϕ. An example that we already mentioned in the introduction is that ∇ of an insulated region. Imagine a thin uniform metal plate that is insulated so no heat can enter or escape. Some time after a given temperature distribution is maintained along the edge of the plate, the temperature distribution inside the plate will reach a steady-state, that will be given by a harmonic function ϕ.
n Throughout this section we write Br(x) R for the ball of radios r centred at x Ω n ⊂ ∈ and Sr(x) R for the corresponding sphere. We set ⊂
cn := vol(S1(0))
n 2 Theorem 4 (Mean value Theorem). Let Ω R and let ϕ C (Ω) be a harmonic ⊂ ∈ function. Then, for each x Ω and r > 0 such that Br(x) Ω one has ∈ ⊂ 1 Z ϕ(x) = n−1 ϕ(y) dσ(y). (2.2) r cn Sr(x)
Proof. Without loss of generality assume that x = 0. By Green’s identities, for any n 1 U R open and φ, ψ C (U), ⊂ ∈ Z Z (φ∆ψ ψ∆φ) dx = (ψ∂νφ φ∂νψ) dσ. (2.3) U − ∂U −
2−n For 0 < ε < r set U := Br(0) Bε(0). Define φ := ϕ and ψ(y) := y whenever \ | | n = 2 and ψ(y) = log( y ) when n = 2. Let us treat the case n = 2 (the other is done 6 | | 6 similarly). It turns out that ∆ψ = 0 on Ω and ∂ψ ∂ψ = (2 n)r1−n and = (2 n)ε1−n. ∂ν |Sr(0) − ∂ν |Sε(0) − − Then, Z 0 = (ϕ∆ψ ψ∆ϕ) dx Ω − Z Z = (ψ∂νϕ ϕ∂νψ) dσ (ψ∂νϕ ϕ∂νψ) dσ Sr(0) − − Sε(0) − Z Z 2−n 2−n = r ∂νϕ dσ ε ∂νϕ dσ Sr(0) − Sε(0) Z Z (2 n)r1−n ϕ dσ + (2 n)ε1−n ϕ dσ − − Sr(0) − Sε(0)
It is also clear from (2.3) that if we pick U := Br(0) or U := Bε(0), ψ := 1 and φ := ϕ, then Z Z ∂νϕ dσ = 0 and ∂νϕ dσ = 0. Sr(0) Sε(0) 30 Laplacian in Euclidean spaces
Therefore, Z Z 0 = r1−n ϕ dσ + ε1−n ϕ dσ. − Sr(0) Sε(0)
By continuity of ϕ,
1 Z 1 Z ϕ(y) dσ(y) = ϕ(y) dσ(y) ϕ(0). n−1 n−1 ε→0 r cn Sr(0) ε cn Sε(0) −→
A function ϕ that satisfies (2.2) is said to have the mean value property. Satisfying the mean value property is equivalent to
n Z ϕ(x) = n ϕ(y) dy. (2.4) r cn Br(x)
Indeed, equation (2.4) follows from integrating ϕ(x)rn−1 with respect to r. Equation (2.2) follows from differentiating ϕ(x)rn with respect to r.
In addition, the mean value property is also equivalent to satisfying
1 Z ϕ(x) = ϕ(x + rw) dS(w). (2.5) cn S1(0) where dS is the area measure on the unit sphere. This follows easily from performing the change of variables y = x + rω. Next we prove a converse to the Mean value Theorem.
n Theorem 5. Let Ω R and ϕ C(Ω) satisfy the mean value property. Then ϕ is ⊂ ∈ smooth and harmonic in Ω.
Corollary 6. Harmonic functions are smooth.
∞ Proof of Theorem 5. Let u be a Friederich’s mollifier. That is, u C (B1(0)) is a ∈ 0 radial function satisfying R u(x)dx = 1. For ε > 0 define B1(0)
1 y uε(y) := u . εn ε
We will prove that ϕ(x) = uε ϕ(x) for x Ω with with 0 < ε < dist(x, ∂Ω). Since uε is ∗ ∈ smooth, it will follow that ϕ is smooth. In what follows, since u is radial, we introduce 2.4 Harmonic functions 31
v(r) := u(rw) for r [0, ) and w S1(0). ∈ ∞ ∈ Z uε ϕ(x) = ϕ(y)uε(y x)dy ∗ Ω − Z = ϕ(x + y)uε(y)dy Ω 1 Z y = n ϕ(x + y)u dy ε Bε(0) ε Z = ϕ(x + εy)u(y)dy B1(0) Z 1 Z = ϕ(x + εrw)u(rw)rn−1 dS(w) dr 0 S1(0) Z 1 Z = v(r)rn−1 ϕ(x + εrw) dS(w) dr 0 S1(0) Z 1 n−1 = ϕ(x)cn v(r)r dr 0 = u(x).
It remains to show that ϕ is harmonic. Since ∆ϕ is continuous, we deduce that ∆ϕ = 0 from the fact that for all r > 0 Z Z ∆ϕ(y) dy = = ∂νϕ(y)dS(y) Br(x) − Sr(x) ∂ Z = rn−1 ϕ(x + rw)dS(w) − ∂r S1(0) n−1 ∂ = r (cnϕ(x)) − ∂r = 0.
Harmonic functions are analytic (Exercise).
n 2 1. Fix x R and let R > 0. Show that if ϕ C (BR(x)) C(BR(x)) is harmonic, ∈ ∈ ∩ then n ∂x ϕ(x) ϕ ∞ . | i | ≤ Rk kL (BR(x))
Hint: Prove and use that ∂xi ϕ is harmonic.
2. Let ϕ, x and R as in the previous part. For m N, prove by induction that there ∈ exists a constant C > 0 independent of m, n and R such that
nmCm−1m! ∂αϕ(x) | | ≤ Rm for any multi-index α with α = m. | | n 3. Prove by Taylor expansion that any harmonic function ϕ on Ω R is analytic. ⊂ 32 Laplacian in Euclidean spaces
n Theorem 7 (The Maximum Principle). Assume Ω R is connected and open. If ϕ ⊂ is harmonic and real-valued on Ω, then
either ϕ(x) < sup ϕ x Ω, or ϕ = sup ϕ. Ω ∀ ∈ Ω
Proof. Consider the set := x Ω: ϕ(x) = sup ϕ . A { ∈ Ω } The set is clearly closed in . The set is also open. Indeed, if ϕ(x) = sup ϕ then A A A Ω ϕ(y) = supΩ ϕ for all y in a ball centred at x for otherwise the Mean Value Theorem would lead to a contradiction. Since is both open and closed in Ω we conclude that A = Ω or = . A A ∅ n Theorem 8. Suppose Ω R is open, and Ω is compact. If ϕ is harmonic and real- ⊂ valued on Ω, and continuous on Ω, then the maximum value of ϕ is achieved on ∂Ω.
Proof. If the maximum is achieved at an interior point, the ϕ must be constant on the connected component of Ω that contains such point, and therefore the maximum is also achieved at the border.
Theorem 9. Suppose Ω is compact and that ϕ, ψ are harmonic on Ω and continuous up to ∂Ω. If ϕ ∂Ω = ψ ∂Ω, then | | ϕ = ψ on Ω.
Proof. Consider the functions φ1 = ϕ ψ and φ2 = ψ ϕ. Both of them are harmonic − − and equal to zero when restricted to the boundary. The result follows from the fact that their maximums are achieved a the boundary.
Dirichlet energy method. Fix f C(∂Ω) and consider the space ∈ 2 = φ C (Ω) : φ ∂Ω = f B { ∈ | } and define th Dirichlet’s energy of φ
1 Z E(φ) = φ(x) 2dx. 2 Ω |∇ |
Theorem 10 (Dirichlet’s Principle). Let Ω be open and bounded. Consider the problem ( ∆φ = 0 in Ω ( ) ∗ φ = f on ∂Ω.
The function ϕ is a solution of ( ) if and only if ∈ B ∗ E(ϕ) = min E(φ). φ∈B 2.4 Harmonic functions 33
Proof. Suppose ϕ solves ( ) and let ψ . Then, ∗ ∈ B Z 0 = ∆ϕ(ϕ ψ) dx Ω − Z Z = ϕ (ϕ ψ) dx ∂νϕ (ϕ ψ) Ω ∇ · ∇ − − ∂Ω − Z Z = ϕ 2dx ϕ ψdx Ω |∇ | − Ω ∇ · ∇ Z 1 Z = ϕ 2dx + ( (ψ ϕ) 2 ϕ 2 ψ 2)dx Ω |∇ | 2 Ω |∇ − | − |∇ | − |∇ | 1 Z 1 Z ϕ 2dx ψ 2dx. ≥ 2 Ω |∇ | − 2 Ω |∇ | Therefore, Z Z ϕ 2dx ψ 2dx. Ω |∇ | ≤ Ω |∇ | Since ψ is arbitrary, it follows that ϕ minimizes the energy.
2 Suppose now that ϕ minimizes the energy. Then, for any ψ C0 (Ω) we must have that d ∈ E(ϕ + εψ) ε=0 = 0. Observe that dε | 1 Z E(ϕ + εψ) = ( ϕ 2 + 2ε ϕ ψ + ε2 ψ 2)dx, 2 Ω |∇ | ∇ ∇ |∇ | and therefore, d 1 Z E(ϕ + εψ) = (2 ϕ ψ + 2ε ψ 2)dx. dε 2 Ω ∇ ∇ |∇ | It follows that Z d E(ϕ + εψ) ε=0 = ϕ ψdx dε Ω ∇ ∇ Z Z = ∆ϕ ψ dx + ∂νϕ ψds Ω ∂Ω Z = ∆ϕ ψ dx. Ω
We then must have R ∆ϕ ψ dx = 0 for all ψ C2(Ω), and so ∆ϕ = 0 which implies Ω ∈ 0 that ϕ is harmonic.
CHAPTER 3
A very brief review of differential and Riemannian geometry
3.1 Differentiable Manifolds
A topological manifold is a topological space (E, τ) so that
1. It is Hausdorff. 2. x E there exists (U, ϕ) with U open and x U, such that ϕ : U ϕ(U) ∀ ∈ ∈ → is a homeomorphism. The pair (U, ϕ) is called chart and the real numbers (x1, . . . , xn) = ϕ(x) are called local coordinates. 3.( E, τ) has a countable basis of open sets.
A Ck-differentiable structure on a topological manifold M is a family of charts = (Uα, ϕα so that U { } S 1. Uα = M
−1 k k 2. If Uα Uβ = then ϕβ ϕ : ϕα(Uα Uβ) ϕβ(Uα Uβ) are C with C inverse. ∩ 6 ∅ ◦ α ∩ → ∩ In this case we say that (Uα, ϕα) and (Uβ, ϕβ) are compatible.
3. Completness property: If (V, ψ) is a chart which is compatible with every (Uα, ϕα) ∈ then (V, ψ) . U ∈ U n n n Examples of differential manifolds include R , the sphere S , the torus T . Products of manifolds are manifolds as well.
k Let M be a manifold and W M open. We say that f : W R is a C -differentiable ⊂ → map if for all x W there exists (U, ϕ) coordinate chart (with x U) so that −1 ∈ k ∈ f ϕ : ϕ(W U) R is C . ◦ ∩ → 36 A very brief review of differential and Riemannian geometry
∞ ∞ Let A M. We say that f : A R is C if it has a C extension to an open set ⊂ → U M such that A U. ⊂ ⊂ Let M and N be differentiable Ck- manifolds. We will say that f : M N is → a C`-differentiable map, ` k, if forall x M there exists (U, ϕ) coordinate ≤ ∈ chart with x U and (V, ψ) coordinate with f(x) V so that f(U) V and −1 ∈ ` ∈ ⊂ ψ f ϕ : ϕ(W U) R is C -differentiable as a map of euclidean spaces. ◦ ◦ ∩ → A manifold with boundary is a Hausdorff space with a countable basis of open sets and a differentiable structure (Uα, ϕα): α A such that it has compatibility on over- {n ∈ } laps and ϕα(Uα) is open in H = (x1, . . . , xn): x1 > 0 . We denote the boundary of { } M by ∂M.
Examples of a manifold with boundary include intervals (a, b), (a, b], [a, b], balls, open n subsets of R with some (or none) pieces of its boundary attached, open subsets of a manifold. Let f, g be C∞ in a neighborhood of x M. We say that f g if there exists U M ∈ ∼ ∞ ⊂ open so that f(y) = g(y) for all y U. The class [f]x is called germ of C function ∈ ∞ at x. The set of germs at x is denoted by C (x, R).
∞ Let x M and Xx : C (x, R) R. If for every chart (U, ϕ) about x we have that ∈ → there exist a1, . . . , an R so that ∈ n X ∂(f ϕ−1) Xx([f]x) = ai ◦ (ϕ(x)), ∂x i=1 i we will say that Xx is a tangent vector at x . If the equation holds for some chart (U, ϕ) about x, then it holds for every C∞-compatible chart overlapping at x.
The tangent space to M at x is the vector space of tangent vectors based at x. We will denote it by TxM. If n = dim M then dim TxM = n. n o Given a coordinate chart (U, ϕ) about x M the basis ∂ , 1 i n is called ∈ ∂xi x ≤ ≤ the natural basis of TxM associated to this chart, where
∂ ∂f ϕ−1 [f]x := ◦ (ϕ(x)). ∂xi x ∂xi
Given f : M N a C∞-map and x M we define the push-forward of f at x as → ∈ follows: (f∗)x : TxM T N (f∗)xXx([g] ) := Xx([g f]x). → f(x) f(x) ◦ The push-forward is what we know as the differential of a function. For this reason we sometimes write (f∗)x = dxf. 3.1 Differentiable Manifolds 37
With this definition we have ∂ −1 ∂ = ϕ∗ . ∂xi x ∂xi ϕ(x) Let ϕ and ψ be charts about x. Let
−1 ψ ϕ (x1, . . . , xn) = (y1(x1, . . . , xn), . . . , yn(x1, . . . , xn)). ◦ Then ∂ X ∂yk ∂ = (ϕ(x)) . ∂xi x ∂xi ∂yk x k We define the cotangent space to x at M to be the dual space of T M. We denote n o x it by T ∗M. The dual basis of ( ∂ ) , 1 i n is denoted by (dx ) , 1 i n x ∂xi x i x ≤∗ ≤ { ≤ ≤ } and it is known as the natural basis of Tx M.
Let ϕ and ψ be charts about x. Let
−1 ϕ ψ (y1, . . . , yn) = (x1(y1, . . . , yn), . . . , xn(y1, . . . , yn)). ◦ Then X ∂xi (dxi)x = (ψ(x))(dyk)x. ∂yk k The tangent bundle of M, denoted by TM is defined as follows: a TM := TxM. x∈M The tangent bundle TM is equipped with a projection map π : TM M defined as → π(Xx) = x for Xx TxM. ∈ A vector field is a section of the tangent bundle. That is, a map X : M TM → that satisfies that π X(x) = X(x). ◦ To each point x M the vector field X assigns a tangent vector at x, X(x) TxM. k ∈ ∈ The space of C -vector fields will be denoted by ΓCk (TM).
Properties of vector fields: 1. X(f + g) = X(f) + X(g)
2. X(λf) = λX(f) λ R ∀ ∈ 3. X(fg) = gX(f) + fX(g) The cotangent bundle of M, denoted by T ∗M is defined as follows:
∗ a ∗ T M := Tx M. x∈M 38 A very brief review of differential and Riemannian geometry
3.2 Differential forms
Let V be a real n-dimensional space and let V ∗ be its dual space. We define the space of alternating k-forms as follows: k ∗ Λ (V ) = ω : V V (k times) R : ω is linear and alternating . { ⊕ · · · ⊕ → } k ∗ n!k! Observe that dim Λ (V ) = (n−k)! . The form ω is linear and alternating if ω(v1, . . . , vn) is linear in each argument and π ω(v , . . . , v ) = ( 1) ω(v1, . . . , vn). π(1) π(k) − k ∗ ` k ∗ Λ (T M) := x∈M Λ (Tx M). Choose a chart (U, ϕ) about x with local coordinates k ∗ (x1, . . . , xn). An element ωx Λ (T M) is called k-form at x and can be written as ∈ x X ωx = ai ...i (dxi )x (dxi )x. 1 k 1 ∧ · · · ∧ k 1≤i1<··· k ∗ n!k! Λ (T M) is a manifold of dimension n + (n−k)! . We define a k-form on M as a section of the bundle π :Λk(T ∗M) M. That is, a C∞ k ∗ → map ω : M Λ (T M) so that π ω = idM . We denote the space of k-forms on M by k → ∗ Ln ◦ k 0 ∞ Ω (M). We write Ω (M) := Ω (M) and Ω (M) = C (M, R). Let f : M N k=0 → be a C∞ map. We define the pull back of f as the map f ∗ :Ω∗(N) Ω∗(M) so that → ∗ 0 ∞ 1. f (g) = g f for f Ω (N) = C (N, R). ◦ ∈ ∗ k 2.( f ω)x(X1,...,Xk) = ω (f∗X1, . . . , f∗Xk) for ω Ω (N) with k 1. f(x) ∈ ≥ Properties of the pull-back map. 1. f ∗(ω τ) = f ∗ω f ∗τ ∧ ∧ 2. f ∗(gω + hτ) = f ∗(g) f ∗ω f ∗(h) f ∗τ ∧ 3.( f g)∗ = g∗ f ∗ ◦ ◦ Proposition. Pull-backs and d commute: d(f ∗ω) = f ∗(dω). Integral of n-forms. Let M be an orientable manifold of dimension n. n n 1 n 1. If ω Ω (R ) has compact support, and ω = fdx dx then ∈ ∧ · · · ∧ Z Z ω := fdx1 . . . dxn. Rn Rn 2. If ω Ωn(M) we define ∈ Z Z Z X X ∗ −1 ω = ραω := (ϕα) (ραω) [M] α∈a Uα α∈a ϕ(Uα) where (Uα, ϕα): α A is a positively oriented atlas and ρα : α A is a { ∈ } { ∈ } partition of unity subordinate to (Uα, ϕα): α A . { ∈ } 3.3 Riemannian Geometry 39 Theorem 11 (Stokes Theorem). Let M be a compact differentiable manifold of dimen- sion n with boundary ∂M. Let ω Ωn−1(M). Then, ∈ Z Z d ω = ω. M ∂M 3.3 Riemannian Geometry A Riemannian Manifold is a pair (M, g) where M is a C∞ manifold and g is a map that assigns to any x M a non-degenerate symmetric positive definite bilinear form ∈ , : TxM TxM R such that for all X,Y ΓC∞ (TM), the map h· ·ig(x) × → ∈ x X(x),Y (x) 7→ h ig(x) is smooth. Notation. Let (U, ϕ) be a chart with local coordinates (x1, . . . , xn), and consider the ∂ ∂ corresponding natural basis ( 1 )x,..., ( n )x of TxM. We adopt the following no- { ∂x ∂x } tation: ∂ ∂ gij(x) := i , j . ∂x x ∂x x g(x) ij We will also denote by g the entries of the inverse matrix of (gij)ij. Proposition. Let M and N be manifolds and let g be a Riemannian metric on N. Let f : M N be a C∞ inmersion. Then the map f ∗ defined below defines a Riemannian → metric f ∗g on M: X(x),Y (x) ∗ := f∗X(x), f∗Y (x) . h i(f g)(x) h ig(x) Theorem. Every manifold carries a Riemannian metric. n n n n Examples of Riemannian manifolds are Ω R , H , S and T . ⊂ Given two Riemannian manifolds (M, gM ) and (N, gN ) we say that a map Φ : (M, gM ) → (N, gN ) is a local isometry provided ∗ Φ (gN ) = gM . If Φ is a local isometry and a diffeomorphism at the same time, then we say that Φ is an isometry. 2 L - Integrable vector fields. Given any two vector fields X,Y ΓC∞ (TM) the ∈ function on M x X(x),Y (x) is smooth and real valued. Therefore, we may 7→ h ig(x) define an inner product on ΓC∞ (TM) by Z X,Y g := X(x),Y (x) g(x) ωg(x). h i M h i 40 A very brief review of differential and Riemannian geometry The completion of ΓC∞ (TM) with respect to ( , )g is a Hilbert space denoted by · · ΓL2 (TM). Geodesic normal coordinates. For (x, v) TM write γx,v for the geodesic on M ∈ starting at x with velocity v. If the geodesic γx,v(t) is defined on the interval ( δ, δ), then the geodesic γx,av, a R, − ∈ a > 0, is defined on the interval ( δ/a, δ/a) and γx,av(t) = γx,v(at). − In addition, given x M, there exist a neighborhood V of x in M, a number ε > 0 and ∈ a C∞ mapping γ :( 2, 2) M, − × U → = (y, w) TM; y V, w TM , w < ε , U { ∈ ∈ ∈ | | } such that t γy,w(t), t ( 2, 2), is the unique geodesic of M which, at the instant → ∈ − t = 0, passes through y with velocity w, for every y V and for every w TqM, with ∈ ∈ w < ε. | | The exponential map exp : M is defined by U → v exp(y, v) = γ(1, y, v) = γ v , y, , (y, v) . | | v ∈ U | | In most applications we shall utilize the restriction of exp to an open subset of the tangent space TxM, that is, exp : Bε(0) TyM M, y ⊂ → expy(v) = exp(y, v). Given x M there exists ε > 0 such that the exponential map exp : Bε(0) TxM ∈ x ⊂ → M is a diffeomorphism. Taking an orthonormal basis v1 . . . , vn of TxM one can define a diffeomorphism n Bε(0) R Bε(0) M ⊂ −→ ⊂n X (x1, . . . , xn) exp xi vi . 7→ x i=1 The coordinate map y (x1(y), . . . , xn(y)) is called geodesic normal coordinates. 7→ CHAPTER 4 The Laplacian on a Riemannian manifold 4.1 Definition Gradient. Given a Riemannian manifold (M n, g) and a function ϕ C∞(M), the ∈ differential map dxϕ : TxM R is linear for all x M. Thus, there exists a vector → ∈ field on TM named gradient of f and denoted by gϕ so that ∇ gϕ(x),Xx = dxϕ(Xx) for all Xx TxM. h∇ ig(x) ∈ Formally speaking, here is the definition. Definition 12. The gradient is the operator ∞ g : C (M) ΓC∞ (TM) ∇ → making gϕ, X g = dϕ(X) for all X ΓC∞ (TM). h∇ i ∈ Let’s find the expression for the gradient vector field in local coordinates (x1, . . . , xn). It has to be a linear combination of the form ϕ = Pn a ∂ for some coefficients g i=1 i ∂xi ∞ ∂ ∂ϕ Pn ∇ ∂ϕ Pn ij ∂ϕ ai C (M). Since dϕ( ) = we get amgmi = and so aj = g ∈ ∂xi ∂xi m=1 ∂xi i=1 ∂xi which implies n X ij ∂ϕ ∂ gϕ = g . ∇ ∂x ∂x i,j=1 i j Since d(ϕ + ψ) = dϕ + dψ for all ϕ, ψ C1(M) we get ∈ g(ϕ + ψ) = gϕ + gψ. ∇ ∇ ∇ Furthermore, since d(ϕ ψ) = ϕ dψ + ψ dϕ we have · · · g(ϕ ψ) = ϕ gψ + ψ gϕ. ∇ · · ∇ · ∇ 42 The Laplacian on a Riemannian manifold Divergence. Given a n-form ω Ωn(M) where n =dim(M), and any vector field X ∈ on M one can define the (n 1)-form ι ω Ωn−1 by − X ∈ ιX ω(X1,...,Xn−1) = ω(X,X1,...,Xn−1) where X1,...,Xn−1 are any vector fields on M. Since d(ιX ω) is an n-form we know there must exist a number divωX making d(ι ω) = divωX ω. X · If ωg is the volume form of (M, g), then the number divgX := divωg X is known as the divergence of X. Formally speaking we have the following definition. Definition 13. The divergence is the operator ∞ divg :ΓC∞ (TM) C (M) → making d(ι ωg) = divgX ωg for all X ΓC∞ (TM). X · ∈ We will now find the expression for divg in local coordinates (x1, . . . , xn). For X = Pn ∂ bj ΓC∞ (TM) we get j=1 ∂xj ∈ ! ! ∂ ∂ˆ ∂ ∂ ∂ˆ ∂ ιX ωg ,..., ,..., = ωg X, ,..., ,..., ∂x1 ∂xi ∂xn ∂x1 ∂xi ∂xn i−1 ∂ ∂ = ( 1) ωg ,...,X,..., − ∂x1 ∂xn i−1p ∂ ∂ = ( 1) det g dx1 dxn ,...,X,..., − | | ∧ · · · ∧ ∂x1 ∂xn i−1p = bi( 1) det g . − | | Therefore, n ! X i−1p d(ι ωg) = d bi( 1) det g dx1 dxˆ i dxn X − | | ∧ · · · ∧ ∧ · · · ∧ i=1 n X i−1 ∂ p = ( 1) (bi det g ) dxi dx1 dxˆ i dxn − ∂x | | ∧ ∧ · · · ∧ ∧ · · · ∧ i=1 i n X ∂ p = (bi det g )dx1 dxn ∂x | | ∧ · · · ∧ i=1 i n 1 X ∂ p = p (bi det g ) ωg, det g ∂xi | | · | | i=1 Pn ∂ and so, for X = bj ΓC∞ (TM), j=1 ∂xj ∈ n 1 X ∂ p divgX = p (bi det g ). (4.1) det g ∂xi | | | | i=1 4.1 Definition 43 Since for any X,Y Γ(TM) and ω Ωn(M) we have ι ω = ι ω + ι ω it follows ∈ ∈ X+Y X Y that divg(X + Y ) = divgX + divgY. Furthermore, from (4.1), we get that for any ϕ C∞(M) ∈ divg(ϕX) = ϕ divgX + gϕ, X . (4.2) h∇ ig We are now in conditions to define our star operator known as the Laplacian, or Laplace operator, or Laplace-Beltrami operator. Definition 14. The Laplacian on (M, g) is the operator ∞ ∞ ∆g : C (M) C (M) → defined as ∆g = divg g. − ◦ ∇ ∞ Since both g and divg are linear operators it follows that for any ϕ, ψ C (M) ∇ ∈ ∆g(ϕ + ψ) = ∆gϕ + ∆gψ. In addition we have ∆g(ϕ ψ) = ψ∆gϕ + ϕ∆gψ 2 gϕ, gψ . · − h∇ ∇ ig In order to prove the last equality we observe that from (4.2) we have divg(ψ gϕ) = ψ ∆gϕ + gϕ, gψ ∇ − h∇ ∇ ig and so ∆g(ϕ ψ) = divg g(ϕ ψ) · − ∇ · = divg(ψ gϕ) div(ϕ gψ) − ∇ − ∇ = ψ∆gϕ + ϕ∆gψ 2 gϕ, gψ . − h∇ ∇ ig Laplacian in local coordinates. From the expression of g and divg in local coordi- ∇ nates (x1, . . . , xn) it is straightforward to see that n 1 X ∂ ijp ∂ ∆g = p g det g . − det g ∂xi | |∂xj | | ij=1 It is worth mentioning that from all the local expressions it follows that if ϕ Ck(M) k−1 k−2 ∈ then gϕ C (M) and so ∆gϕ = divg gϕ C (M). ∇ ∈ − ∇ ∈ Average over orthogonal geodesics (local definition). Given x M there exists ∈ ε > 0 such that the exponential map exp : Bε(0) TxM M is a diffeomorphism. x ⊂ → Taking an orthonormal basis v1 . . . , vn of TxM one can define a diffeomorphism n Bε(0) R Bε(0) M ⊂ −→ ⊂n X (x1, . . . , xn) exp xi vi . 7→ x i=1 44 The Laplacian on a Riemannian manifold The coordinate map y (x1(y), . . . , xn(y)) gives rise to the so called geodesic normal 7→ coordinates. For x M consider an orthonormal basis v1, . . . vn of TxM and write (x1, . . . , xn) for ∈ the geodesic normal coordinates system around x determined by such basis. Since in ∂gij these coordinates gij(x) = δij and (x) = 0 for all i, j, k = 1, . . . , n, it follows that ∂xk n X ∂2ϕ ∆gϕ(x) = (x). − ∂x2 i=1 i For each i = 1, . . . , n let γi be the geodesic satisfying the conditions γi(0) = x and γ˙i(0) = v1. Since for any i, j = 1, . . . , n we have xi(γj(t)) = δij t, then 2 2 ∂ ϕ d 2 (x) = 2 ϕ(γi(t)) . ∂xi dt t=0 It follows that n X d2 ∆gϕ(x) = ϕ(γi(t)) . − dt2 t=0 i=1 Alternative definitions. - The Laplacian is sometimes defined as ∆g = traceg(Hessg) where Hessg is the Hes- − sian operator on (M, g). - If we write δg for the adjoint operator of d, then ∆g = δg d. This definition can be ∗ ∗ generalized to define a Laplace operator acting on forms. Indeed, ∆g :Ω (M) Ω (M) → is defined as ∆g := δgd + dδg. P ij ∂2 - Since ∆g = g +lower order terms, the Laplacian is also characterized as − ij ∂xi∂xj the only symmetric linear partial differential operator whose principal symbol is 2. −k · kg 4.2 Examples n Laplacian on R . n n Let g n be the euclidean metric on R . Since gij(x) = δij for all x R and i, j = R ∈ 1, . . . , n it follows that n X ∂2 ∆g n = . R − ∂x2 i=1 i Laplacian on H. 2 Consider the upper half plane H = (x, y) R : y > 0 endowed with the hyperbolic { ∈ } metric 1 ! y2 0 gH(x, y) = 1 . 0 y2 4.3 The Laplacian under conformal deformations (Exercise) 45 It is straightforward that 2 2 2 ∂ 2 ∂ ∆g = y y . H − ∂x2 − ∂y2 Laplacian on S2. 2 Let gS2 be the round metric on the 2-sphere S . Endow the sphere with spherical coordinates 2 3 T : (0, π) (0, 2π) S R × → ⊂ T (θ, φ) = (sin θ cos φ, sin θ sin φ, cos θ). Since 1 0 g 2 (θ, φ) = , S 0 sin2 θ we get: 1 ∂ θθp ∂ ∂ φφp ∂ ∆ = g det g 2 + g det g 2 gS2 p S S − det g 2 ∂θ | |∂θ ∂φ | |∂φ | S | 1 ∂ ∂ ∂ 1 ∂ = sin θ + −sin θ ∂θ ∂θ ∂φ sin θ ∂φ 1 ∂ ∂ 1 ∂2 = sin θ . −sin θ ∂θ ∂θ − sin2 θ ∂φ2 And so, in spherical coordinates, 1 ∂ ∂ 1 ∂2 ∆g = sin θ . (4.3) S2 −sin θ ∂θ ∂θ − sin2 θ ∂φ2 4.3 The Laplacian under conformal deformations (Exer- cise) Consider a conformal deformationg ˜ of the metric g. Thas is,g ˜ = ef g with f C∞(M). ∈ When you modify a metric conformally all you are doing is to change the distances between points while maintaining the angles between vectors. The aim of this exercise is to prove that −f n −2f ∆g˜ = e ∆g + 1 e gf. − 2 ∇ −f Notice that on surfaces this formula simplifies to ∆g˜ = e ∆g. Operators that satisfy such law are known as conformally covariant operators. We suggest you prove the assertion by showing the following: −f 1. g˜ = e g. ∇ ∇ n −f 2. divg˜(X) = divg(X) + 2 e X(f). −f n −2f 3.∆ g˜ = e ∆g + 1 e gf. − 2 ∇ 46 The Laplacian on a Riemannian manifold 4.4 The Laplacian commutes with isometries The purpose of this exercise is to show that the Laplace operator commutes with isome- tries. That is, given any two manifolds (M, gM ), (N, gN ) and an isometry Φ : (M, gM ) → (N, gN ) between them, show that ∗ ∗ ∆gM Φ = Φ ∆gN . To simplify your approach to this problem we suggest you break its solution into three steps. Using the definitions of g and divg show: ∇ ∗ 1.Φ ∗ g Φ = g . ∇ M ∇ N ∗ 2.Φ ∗ divgN Φ = divgM . ∗ ∗ 3.∆ gM Φ = Φ ∆gN . Solutions: 1. Let ϕ C1(N) and V a vector field on TN. Then ∈ ∗ ∗ −1 Φ∗ gM Φ (ϕ),V = Φ∗ gM Φ (ϕ), Φ∗Φ∗ V h ∇ igN ∇ gN ∗ −1 = gM Φ (ϕ), Φ∗ V ∇ gM ∗ −1 = d(Φ ϕ)[Φ∗ V ] −1 = ϕ∗Φ∗[Φ∗ V ] = dϕ(V ) = g ϕ, V . h∇ N igN 2. If we fix any X Γ(TM), the assertion would follow if could prove d(ιX ωgM ) = ∗ ∈ ∗ −1 Φ divgN Φ∗(X) ωgM . But this is equivalent to showing that (Φ ) (dιX ωgM ) = · ∗ −1 divg Φ∗(X) (Φ ) ωg , or what is the same, we need to show that N · M ∗ −1 d((Φ ) (ι ωg ) = divg (Φ∗X) ωg . X M N · N ∗ −1 ∗ −1 This last equality is true since (Φ ) (ιX ωgM ) = ιΦ∗X (Φ ) (ωgM ). 3. ∗ ∗ −1 ∗ ∗ ∆g Φ = divg g Φ = divg (Φ ) Φ g Φ M M ∇ M M ∗ ∇ M −1 ∗ ∗ −1 −1 = divgM (Φ∗) gN = Φ (Φ ) divgM (Φ∗) gN ∗ ∇ ∗ ∇ = Φ divg g = Φ ∆g N ∇ N N 4.5 The Laplacian and Riemannian coverings Given two Riemannian manifolds (M, gM ) and (M,˜ g ˜ ), a map p : M˜ M is a M → Riemannian covering if p is a differentiable covering (that is, a local surjective home- ∗ omorphism) that is a local isometry. In particular, p gM = g ˜ . If p : M˜ M is a M → 4.6 The Laplacian on product manifolds 47 Riemannian covering, the deck transformation group of the covering p is the group of homeomorphisms α : M˜ M˜ satisfying → p α = p. ◦ When p : M˜ M is a Riemannian covering, the elements of the deck transformation → group are isometries. Indeed, they are homeomorphisms and local isometries: ∗ ∗ ∗ ∗ ∗ α g ˜ = α p gM = (p α) gM = p gM = g ˜ . M ◦ ◦ M If p : M˜ M is a Riemannian covering, the functions on M can be identified with that → of M˜ that are invariant under the desk transformation group. Indeed, ifϕ ˜ : M C → satisfiesϕ ˜ α =ϕ ˜ for every deck transformation α, then there exists a unique function ◦ ϕ : M C for whichϕ ˜ = ϕ p. In particular, the eigenvalues of M are precisely → ◦ those of M˜ in whose eigenspace there are eigenfunctions of M˜ invariant under the deck transformation group. Indeed, ifϕ ˜ is an eigenfunction of ∆gM˜ of eigenvalue λ invariant under the deck transformation group, then there exists a unique ϕ : M R such that → λϕ p = λϕ˜. On the other hand, ◦ ∗ ∗ λϕ˜ = ∆g ϕ˜ = ∆g p ϕ = (p ∆g )ϕ = ∆g ϕ p M˜ M˜ M M ◦ where we used that p is an isometry and that the Laplacian commutes with it. It follows that we must have ∆gϕ = λϕ. In addition, if ϕ is an eigenfunction of ∆g , then ϕ p is an eigenfunction of ∆g˜ . M ◦ M Indeed, ∗ ∗ ∆g (ϕ p) = p ∆g ϕ = p (λϕ) = λ ϕ p. M˜ ◦ M ◦ Quotients by discrete group of isometries. A discrete group Γ is said to act properly on a Riemannian manifold (M,˜ g ˜ ) if for anyx, ˜ y˜ M˜ there exist open M ∈ neighborhoods U and V ofx ˜ andy ˜ respectively such that # γ Γ: γU V = < . { ∈ ∩ 6 } ∞ The group Γ is said to act freely if for any γ1, γ2 Γ for which there exists a point ∈ x˜ M˜ one has γ1 = γ2. The following theorem gives the existence of a canonical metric ∈ on a manifold obtained as a quotient by a discrete group of isometries. Theorem 15. Let Γ be a discrete group of isometries acting properly and freely on a ˜ Riemannian manifold (M, gM˜ ). There exists a unique canonical Riemannian metric gM on the quotient M = M/˜ Γ such that p :(M,˜ g ˜ ) (M, gM ) is a Riemannian covering. M → Examples of manifolds where we may apply this result are the circle, the torus and the Klein bottle. 4.6 The Laplacian on product manifolds If (M, g) and (N, h) are Riemannian manifolds, we can endow M N with the product × metric: If πM : M N M and πN : M N M are the projections then the × → × → product metric is defined as follows X,Y g⊕h(x ,x ) := X,Y π∗ (g)(x ) + X,Y π∗ (g)(x ) h i 1 2 h i M 1 h i N 2 48 The Laplacian on a Riemannian manifold where (x1, x2) M N. ∈ × Note that by the Stone-Weirstass Theorem the set of functions (x1, x2) φ(x1)ψ(x2) 7→ ∈ C∞(M N) with φ C∞(M) and ψ C∞(N) is dense in L2(M N). In addition, if × ∈ ∈ × ∆gM φ = λφ and ∆gN ψ = βψ, then ∆g ⊕g (φ ψ) = ψ∆g ⊕g φ + φ∆g ⊕g ψ 2 g ⊕g φ, g ⊕g ψ g ⊕g M N · M N M N − h∇ M N ∇ M N i M N = ψ∆gM φ + φ∆gN ψ = (λ + β)φ ψ. · As we will show later, if M and N are compact, one can find an orthonormal basis φj j 2 2 { } of L (M, gM ) (reps. ψj j of L (N, gN )) of eigenfunctions of ∆g with eigenvalues λj { } M (resp. of eigenfunctions of ∆gN with eigenvalues βj). It then follows that ∞ (x1, x2) φj(x1) ψk(x2) j,k C (M N) { 7→ · } ⊂ × is a basis of eigenfunctions of L2(M N) with eigenvalues × λj + βk j,k. { } 4.7 Green’s theorem Theorem 16 (Divergence theorem). Let M be a Riemannian manifold and let X ∈ ΓC1 (TM). Then, Z Z divgX ωg = X, ν σg M ∂M h i where ν is the unit vector normal to ∂M. Proof. We claim that ι ωg = X, ν σg. (4.4) X h i Assuming this holds, since divgX ωg = d(ιX ωg), we get Z Z Z Z divgX ωg = d(ιX ωg) = ιX ωg = X, ν σg M M ∂M ∂M h i as desired. To prove (4.4) let us first show that σg = ιν ωg. Indeed, the metric h on ∂M is defined as follows. Let x ∂M and let v1, . . . , vn ⊥∈ be an orthonormal basis of TxM with v1 = ν (Tx∂M) . Let x1, . . . , xn be the ∈ corresponding system of coordinates on M. Writing h for the metric on ∂M we have D E D E ∂ , ∂ = ∂ , ∂ for all i, j = 2, . . . , n. Note that ∂xi ∂xj h ∂xi ∂xj g ∂ ∂ ∂ ∂ p ∂ ∂ ιν ωg ,..., = ωg ν, ,..., = det h = σg ,..., . ∂x2 ∂xn ∂x2 ∂xn | | ∂x2 ∂xn 4.7 Green’s theorem 49 To prove equation (4.4) we also need to show that ι ωg = X, ν ινωg. But this follows X h ig from the following chain of equalities: ∂ ∂ ∂ ∂ ιX ωg ,..., = ωg X, ,..., ∂x2 ∂xn ∂x2 ∂xn ∂ ∂ = ωg X, ν ν, ,..., h i ∂x2 ∂xn ∂ ∂ = X, ν ινωg ,..., . h i ∂x2 ∂xn Theorem 17 (Green’s Theorem). Let (M, g) be a compact Riemannian manifold with boundary ∂M. Let ψ C1(M) and φ C2(M). Then, ∈ ∈ Z Z Z ∂φ ψ ∆gφ ωg = gψ, gφ ωg ψ σg . M · M h∇ ∇ i − ∂M · ∂ν Proof. From equation (4.2) and Green’s Theorem we have Z Z Z ψ ∆gφ gψ, gφ ωg = divg(ψ gφ) ωg M · − M h∇ ∇ i − M ∇ Z = ψ gφ, ν g σg − ∂M h ∇ i Z ∂φ = ψ σg. − ∂M · ∂ν Corollary 18. If (M, g) is a compact Riemannian manifold without a boundary, then ψ, ∆gφ = gψ, gφ g. h ig h∇ ∇ i Corollary 19 (Formal self-adjointness). If (M, g) is a compact Riemannian manifold without a boundary, then ψ, ∆gφ = φ, ∆gψ . h ig h ig Corollary 20 (Positivity). If (M, g) is a compact Riemannian manifold without a boundary, then ∆gφ, φ 0. h ig ≥ Remark 21. All the previous corollaries are valid for a manifold (M, g) with a boundary where Dirichlet or Neumann boundary conditions are imposed. CHAPTER 5 Examples on manifolds 5.1 Circle Consider the circle T = R/(2πZ). The deck transformation group consists of the trans- lations αj(t) = t + 2πj with j Z. Then, the eigenfunctions of T are eigenfunctions of d2 ∈ 2 on R that satisfy ϕ(t) = ϕ(t + 2πj) for all j Z and all t R. The eigenfunctions − dt ∈ ∈ are therefore ikt t e with k Z. 7→ ∈ Taking real and imaginary parts we get the functions 1, cos(t), sin(t), cos(2t), sin(2t),..., cos(kt), sin(kt),... with eigenvalues 0, 1, 1, 4, 4, . . . , k2, k2,... respectively. 5.2 Torus n Let Γ be an n-dimensional lattice in R . That is, there exist γ1, . . . , γn so that Γ is n generated by γ1e1, . . . , γnen over Z where e1, . . . , en is a basis of R . Consider the { } { } torus n T := R /Γ. We define the dual lattice ∗ n Γ := x R : x, γ Z for all γ Γ . { ∈ h i ∈ ∈ } n n n ∗ n n Let A : R R be so that A(Z ) = Γ. Then, vol(T) = det A. Consider A : R R . ∗ −→1 n ∗ → Then (A ) (Z ) = Γ . Therefore, ∗ ∗ −1 1 vol(T ) = det((A ) ) = . det A 52 Examples on manifolds n Since the eigenfunctions of T are those of R which are invariant under the deck trans- formation group αγ(x) = x + γ : γ Γ we consider the family of functions { ∈ } 2πihx,yi ∗ ϕy(x) := e for y Γ . ∈ Clearly, ϕy is invariant under the deck transformation group. If you want, you may take their real and imaginary part to get the eigenfunctions 1, sin(2πi x, y ), cos(2πi x, y ), y Γ∗ h i h i ∈ with eigenvalues 0, 4π2 y 2, 4π2 y 2, y Γ∗. | | | | ∈ Let us see that the ∗ ϕy : y Γ { ∈ } 2 is a basis of L (T). We first check that these functions are linearly independent by induction: suppose Pk+1 ϕy1 , . . . , ϕyk are linearly independent and suppose also that i=1 aiϕyi = 0. Since Pk+1 Pk ϕyj ϕyi = ϕyj +yi we know that 0 = i=1 aiϕyi−yk+1 = ak+1 + i=1 aiϕyi−yk+1 , after ◦ Pk 2 2 applying the Laplacian we get ai4π yi yk+1 ϕy −y = 0. It then follows that i=1 | − | i k+1 a1 = = ak = 0. Therefore ak+1 = 0. ··· ∗ 2 To see that = span ϕy : y Γ is dense in L (T) we use Stone Weirstarss Theorem. B { ∈ } Clearly is a subalgebra. Let us see that it separates points. Fix two points x1, x2 T B ∈ and assume x1 = x2 (then x1 x2 / Γ). Suppose now that ϕy(x1) = ϕy(x2) for all ∗ 6 2πi−hx −x∈,yi ∗ y Γ . It then follows that e 1 2 = 1 for all y Γ and so x1 x2, y Z for ∈ ∗ ∗ ∗ ∈ h − i ∈ all y Γ which implies that x1 x2 (Γ ) = Γ, and this is a contradiction. ∈ − ∈ Weyl’s law on the Torus We continue to write ωn := vol(B1(0)). Denote by N(λ) the counting function N(λ) := # j : λj < λ . { } Theorem 22 (Weyl’s law on the torus). ωn vol(T) n N(λ) λ 2 λ . ∼ (2π)n → ∞ Note that N(λ) = # j : λj < λ { } = # y Γ∗ : 4π2 y 2 < λ { ∈ | | } = # y Γ∗ : y < √λ/2π { ∈ | | } ∗ √ = # Γ B λ (0) . { ∩ 2π } We therefore define ∗ ∗ N (r) := # Γ Br(0) , { ∩ } 5.3 Rectangular prisms 53 √ ∗ λ and so we have N(λ) = N 2π . We have reduced our problem to show ∗ n ωn n N (r) ωnvol(T)r = r r . ∼ vol(T∗) → ∞ ∗ ∗ Let P (r) denote the number of polygons inside Br(0) and write C (r) for the polygon formed by all the parallelepipeds inside Br(0). C∗(r) r d r − d Observe that P ∗(r) N ∗(r) P ∗(r + d). ≤ ≤ On the one hand, ∗ ∗ ∗ n P (r)vol(T ) = vol(C (r)) vol(Br(0)) = ωnr . ≤ On the other hand, if we write d for the diameter of each parallelogram, we get n ∗ ∗ (r d) ωn P (r)vol(T ) − ≤ ∗ for Br−d(0) C (r). We conclude ⊂ ωn ωn (r d)n P ∗(r) N ∗(r) P ∗(r + d) (r + d)n. vol(T∗) − ≤ ≤ ≤ ≤ vol(T∗) 5.3 Rectangular prisms We work with rectangles Ω = [0, γ1] [0, γn] with Dirichlet boundary conditions. × · · · × π2j2 Since we know that the eigenvalues for [0, γj] are γ2 with j Z+ we know that the j ∈ eigenvalues of Ω are 2 2 2 j1 jn π 2 + + 2 j1, . . . , jn Z+. γ1 ··· γn ∈ 54 Examples on manifolds Weyl’s law on rectangular prisms Consider the prism Ω = [0, γ1] [0, γn] and the lattice Γ generated by γ1e1, . . . , γnen ×· · ·× n ∗ { e1 en } where e1, . . . , en is the standard basis of R . Then Γ is generated by ,..., . { } { γ1 γn } It follows that 2 2 n 2 j1 jn N(λ) = # (j1, . . . , jn) Z+ : π 2 + + 2 < λ ∈ γ1 ··· γn ( ) √λ = y Γ∗, y > 0, y < ∈ | | π ∗ √ # Γ B λ (0) ∩ π ∼ 2n √ ∗ λ N π = 2n √ ∗ λ ωn n/2 ∗ Since N ∗ n λ we get the desired result. Observe that vol(T ) = π ∼ vol(T )π 1 ... 1 = 1 . γ1 γn vol(Ω) Theorem 23 (Weyl’s law on prisms). ωn vol(Ω) n N(λ) λ 2 λ . ∼ (2π)n → ∞ 5.4 Sphere n n+1 Let f(ξ1, . . . , ξn) represent the spherical coordinate system for the sphere S R . n+1 ⊂ Then any x R can be written as rf(ξ1, . . . ξn) for r > 0 and is represented by ∈ n+1 the coordinates (r, ξ1, . . . , ξn). We proceed to compute the euclidean metric of R in n n terms of the round metric on S . Notice that as vector fields in R n+1 n+1 ∂ X ∂(rfi) ∂ X ∂ = = fi , and ∂r ∂r ∂x ∂x i=1 i i=1 i n+1 n+1 ∂ X ∂(rfi) ∂ X ∂fi ∂ = = r . ∂ξ ∂ξ ∂x ∂ξ ∂x j i=1 j i i=1 j i Therefore, n+1 ∂ ∂ X 2 , = fk = 1, ∂r ∂r g n+1 R k=1 n+1 ∂ ∂ X ∂fk , = r f = 0, ∂r ∂ξ k ∂ξ j g n+1 j R k=1 n+1 ∂ ∂ X ∂fk ∂fk ∂ ∂ , = r2 = r2 , . ∂ξi ∂ξj ∂ξi ∂ξj ∂ξi ∂ξj g n+1 gSn R k=1 5.4 Sphere 55 It then follows that 1 0 gRn+1 (r, ξ) = 2 . 0 r gSn (ξ) It is straightforward to check that in these coordinates then 1 ∂ n ∂ 1 ∆g n+1 = r + ∆gSn . R −rn ∂r ∂r r2 Define k = homogeneous polynomials of degree k , P { } k = P k : ∆g n+1 P = 0 , H { ∈ P R } Hk = P Sn : P k . { | ∈ H } Proposition 24. The spaces k and Hk are isomorphic. In addition, H ∞ n Hk Y C (S ) : ∆g n Y = k(n + k 1)Y . ⊂ { ∈ S − } Proof. The idea is to prove that the restriction map k Hk H → P P Sn 7→ | has inverse Hk k → H Y rkY. 7→ Indeed, if Y Hk, then Y = P Sn for some P k. Since P is a homogeneous ∈ | ∈ H polynomial of degree k we have x x x P (x) = P x = x kP = x kY . k k x k k x k k x k k k k k k n Let us see that if Y Hk then Y is an eigenfunction of ∆gSn . Indeed, Y = P S for ∈ |k some P k, and since P is homogenous of degree k, we must have P (r, ξ) = r Y (ξ). ∈ H Then 0 = ∆g n+1 P R 1 ∂ n+k−1 k−2 = Y (kr ) + r ∆g n Y − rn ∂r S k−2 k−2 = k(n + k 1)r Y + r ∆g n Y − − S k−2 = r (∆g n Y k(n + k 1)Y ). S − − Since any Y Hk are restrictions to the sphere of a harmonic polynomial, the space ∈ Hk is known as the space of spherical harmonics of degree k. 56 Examples on manifolds 2 2 Proposition 25. Write r k−2 := r P : P k−2 . Then, P { ∈ P } 2 k = k r k−2. P H ⊕ P P α Proof. We start defining an inner product on k. For Q k and P = aαx k P ∈ P α ∈ P we set X α (P,Q) := aα∂ Q α which we may rewrite as (P,Q) = P (∂)Q. Note that if both α = β = k, then | | | | ( α! if α = β (xα, xβ) = 0 else. In particular X α X β X aαx , aβx = α! aα bα. α β α It follows that ( , ) is a scalar product on k. P 2 2 2 Note that if P k−2 and Q k, and since r = x + + x , ∈ P ∈ P 1 ··· n+1 2 2 (r P,Q) = P (∂)r (∂) Q = P (∂) ∆g n+1 Q = (P, ∆g n+1 Q) = 0. R R 2 This shows that k is the orthogonal complement of r k−2 with respect to ( , ). H 2 P The proof that k = k + r k−2 needs to be added . ♣ P H P ♣ By induction one proves the following corollary. Corollary 26. 2 4 2k 2k = 2k r 2k−2 r 2k−4 . . . r 0, P H ⊕ H ⊕ H ⊕ H 2 4 2k 2k+1 = 2k+1 r 2k−1 r 2k−3 . . . r 1. P H ⊕ H ⊕ H ⊕ H Corollary 27. If P k, then P Sn is a sum of spherical harmonics of degree k. ∈ P | ≤ Corollary 28. (k + n 2)! dim Hk = (2k + n 1) − . − k!(n 1)! − Proof. Since Hk and k are isomorphic, dimHk = dim k. And from Proposition 25, H H dimHk = dim k dim k−2. P − P It only remains to compute the dimension of k. In order to do that note that the α P monomials x with α = k are a basis for k. Therefore, dim k is the number of ways | | P P in which you can form an n + 1-tuple α = (α1, . . . , αn+1) such that α1 + + αn+1 = k. ··· Imagine you want to know in how many way you can arrange k black balls into n + 1 subsets, and that you separate any two subsets using n white balls. α1 α2 α3 α4 α5 5.4 Sphere 57 It follows that the dimension of k is the number of ways in which you can choose from P a total of n + k black balls a subset of n balls and paint them in white. That is, n + k dim k = . P n Theorem 29. ∞ 2 n M L (S ) = Hk. k=1 Proof. Taking r = 1 in the above corollary we get that ∞ ∞ ∞ Hk = k Sn = k Sn . ⊕k=1 ⊕k=1H | ⊕k=1P | ∞ ∞ n Since k ` k+` we get that k=1 k Sn is a subalgebra of C (S ). Note that P ·P ⊂ P ⊕ nP | it separates points. Indeed, if y, z S are different points, y = (y1, . . . , yn+1) and ∈ x = (z1, . . . , zn+1), then there must exist a coordinate for which yj = zj. Therefore we 6 may choose P (x1, . . . , xn+1) := xj 1 which makes P (y) = P (z). By Stone Weirstrass ∞ ∈ P 2 n 6 Theorem we obtain that Hk is dense in L (S ). ⊕k=1 n n+1 We proved that the eigenfunctions of the Laplacian on the sphere S R are restric- ⊂ tions of harmonic polynomials to the sphere. The eigenspaces are Hk with corresponding eigenvalues k(k + n 1) of multiplicities (2k + n 1) (k+n−2)! . − − k!(n−1)! The 2-sphere. The eigenvalues are k(k + 1) with multiplicities 2k + 1 for k N. ∈ Let us parametrize S2 with spherical coordinates (θ, φ) (sin θ cos φ, sin θ sin φ, cos θ). 7→ Let us first find the first degree spherical harmonics. First observe that 1 = span x, y, z 2+1 P { } and dim 1 = = 3. Also 1 = span x, y, z and therefore P 2 H { } H1 = span sin θ cos φ, sin θ sin φ, cos θ { } with corresponding eigenvalue 2 of multiplicity 3. We now find the second degree spherical harmonics. First observe that the second degree 2 2 2 2+2 harmonic polynomials are 2 = span x , y , z , xy, xz, yz and dim 2 = 2 = 6. 2 2 2P 2 { } P Also 2 = span z x , z y , xy, xz, yz and therefore H { − − } 2 2 2 2 2 2 H2 = span cos θ sin θ cos φ, cos θ sin θ sin φ, { − − sin2 θ cos φ sin φ, sin θ cos φ cos θ, sin θ sin φ cos θ } with corresponding eigenvalue 6 of multiplicity 5. 58 Examples on manifolds To find a basis of Hk for general k, let us try to solve Laplace’s equation by separation of variables. Suppose Yk(θ, φ) = Pk(θ)Φk(φ) solves ∆gS2 Yk = k(k + 1)Yk. Then, by the formula for the Laplacian (4.3) in spherical coordinates we have 1 ∂ ∂ 1 ∂2 sin θ Yk(θ, φ) = k(k + 1)Yk(θ, φ) −sin θ ∂θ ∂θ − sin2 θ ∂φ2 which reduces our problem to the following two systems: Φ00(φ) k = m2, −Φk(φ) 2 sin θ d 0 2 k(k + 1) sin θ + sin θ Pk(θ) = m , Pk(θ) dθ imφ where m is a priori any complex number. Then Φk(φ) = e , and, since Φk is 2π- periodic m must be an integer. Changing variables θ t = cos θ we may rewrite the 7→ second equation as 2 2 00 0 m (1 t )P 2tP + k(k + 1) Pk = 0 − k − k − 1 t2 − which is known as the generalized Legendre’s equation and has 2k+1 = dim Hk solutions known as the associated Legendre polynomials P m(t) where k m k. It then k − ≤ ≤ follows that m Hk = span Y : k m k { k − ≤ ≤ } where m imφ m Yk (θ, φ) = e Pk (cos θ). If we restrict ourselves to real eigenfunctions we then have that Indices reverted be- ♣ cause of the picture! ♣ k Hk = span y : k m k { m − ≤ ≤ } where √2 Cm cos(mφ) P m(cos θ) if m > 0, k k k 0 0 ym(θ, φ) = Ck Pk (cos θ) if m = 0, √2 Cm sin( mφ) P −m(cos θ) if m < 0, k − k m m where C are so that y 2 = 1 k k k k s (2k + 1)(k m )! Cm := − | | . k 4π(k + m )! | | 0 Zonal harmonics. When m = 0 the eigenfunctions yk are known as zonal harmonics. They are invariant under rotations that fix the south and north poles. 0 0 0 yk(θ, φ) = Ck Pk (cos θ). Highest weight harmonics. This is the case k = m. Here, yk(θ, φ) = √2 Ck cos(kφ) P k(cos θ) = √2 Ck ( 1)k(2k 1)!! cos(kφ) sin(θ)k. k k k k − − 170 5.4 Sphere 59 The first few spherical harmonics, in both spherical and cartesian coordinates, expand to: Spherical Cartesian 1 1 l 0 y0(µ,¡) , = 0 = r4º r4º 1 3 3 y1° (µ,¡) sin¡sinµ x, 8 = r4º r4º > 0 3 3 l 1 > y (µ,¡) cosµ z, = > 1 = 4º 4º <> r r 3 3 y1(µ,¡) cos¡sinµ y, > 1 > = r4º r4º > :> 2 15 2 15 y2° (µ,¡) sin¡cos¡sin µ xy, 8 = r4º r4º > > 1 15 15 > y° (µ,¡) sin¡sinµ cosµ yz, > 2 = > r4º r4º > > 0 5 2 5 2 l 2 > y (µ,¡) (3cos µ 1) (3z 1), = > 2 = 16º ° 16º ° <> r r 15 15 y1(µ,¡) cos¡sinµ cosµ xz, > 2 > = r4º r8º > > 2 15 2 2 2 15 2 2 > y (µ,¡) (cos ¡ sin ¡)sin µ (x y ). > 2 = 16º ° 32º ° > r r > :> We illustrate the first basis functions in Figure B.1. B.2 Projection and Expansion The spherical harmonics define a complete basis over the sphere. Thus, any real-valued 171 spherical function f may be expanded as a linear combination of the basis functions 0 y0 l 1 m m f (~!) yl (~!) fl , (B.9) = l 0 m l X= X=° m m where the coefficients fl are computed by projecting f onto each basis function yl y 1 f m ym(~!)yf (0~!)d~!.y1 (B.10) 1° l = l 1 1 Z≠4º Just as with the Fourier series, this expansion is exact as long as l goes to infinity; however, this requires an infinite number of coefficients. By limiting the number of bands to l n 1 we retain = ° 2 1 0 1 2 y2° y2° y2 y2 y2 Figure B.1: Plots of the real-valued spherical harmonic basis functions. Green indicates positive values and red indicates negative values. only the frequencies of the function up to some threshold, obtaining an nth order band-limited approximation f˜ of the original function f : n 1 l ˜ ° m m f (~!) yl (~!) fl . (B.11) = l 0 m l X= X=° Low-frequency functions can be well approximated using only a few bands, and as the number of coefficients increases, higher frequency signals can be approximated more accurately. It is often convenient to reformulate the indexing scheme to use a single parameter i l(l 1) m. With this convension it is easy to see that an nth order approximation can be = + + reconstructed using n2 coefficients, n2 1 ° f˜(~!) yi (~!) fi . (B.12) = i 0 X= 60 Examples on manifolds 5.5 Klein Bottle (Exercise) For a > 0 and b > 0 consider the discrete lattice Γa,b generated by ae1, be2 where { } e1 = (1, 0) and e2 = (0, 1). We write Γˆa,b for the group generated by the translations induced by Γa,b and the transformation 2 2 a α : R R , α(x, y) = x + , b y . → 2 − 1. Show that Γˆa,b is a discrete group of isometries that acts properly and freely on 2 R . 2 2. By the previous part we may endow the Klein bottle Ka,b := R /Γa,b with a canonical Riemannian structure gKa,b . Find the eigenvalues of the Laplacian on (Ka,b, gKa,b ). n n 3. Deduce that the flat torus T = R /Z and Ka,b are never isospectral (that is, they cannot have the same eigenvalues). 5.6 Projective space (Exercise) n Consider the sphere S endowed with the round metric gSn . Consider the isometry α : Sn Sn α(x) = x → − and the group of isometries Γ = Id, α . The projective space is defined as the quo- n n { } tient P (R) := S /Γ. The projective space inherites a canonical Riemanian metric gPn(R). Find the eigenvalues n and eigenfunctions of (P (R), gPn(R)). CHAPTER 6 Heat Operator 6.1 Sobolev Spaces ∞ n We start this chapter reviewing some basics on the Fourier transform. For φ C0 (R ) ∈ its Fourier transform is Z φˆ(ξ) := e−ix·ξφ(x)dx Rn where we write (throughout this section) 1 dx := dx1 . . . dxn. (2π)n/2 2 n We need some notation. For φ, ψ L (R ) their convolution is ∈ Z φ ψ(x) = φ(x y)ψ(y)dy. ∗ Rn − n Also, given an n-tuple α = (α1, . . . , αn) Z+ we write ∈ Dα := ( i)|α|∂α1 . . . ∂αn . − x1 xn The following collection of results is well known: Lemma 30. ∞ n 2 The Fourier transform is an isometry on C0 (R ) in the L -norm, and so it • 2 n extends to an isometry of L (R ). (φ ψ)ˆ= φˆ ψˆ. • ∗ · (φψ)ˆ= φˆ ψˆ. • ∗ R ix·ξ φ(x) = n e φˆ(ξ) dξ. • R 62 Heat Operator (Dαφ)ˆ(ξ) = ξα φˆ(ξ). • (xαφ)ˆ(ξ) = Dαφˆ(ξ). • n ∞ For Ω R open with Ω compact, φ C0 (Ω) and s = 0, 1, 2,... set ⊂ ∈ 1 X α 2 2 φ H := D φ . k k s k k2 |α|≤s If φ Hs < we say the φ belongs to the s- Sobolev Space Hs(Ω). Note that H0(Ω) = 2k k ∞ L (Ω). The Sobolev s-norm can be generalized for s R. In order to do this note that ∈ (setting φ 0 on Ωc) ≡ X X φ 2 = Dαφ 2 = (Dαφ)ˆ 2 k kHs k k2 k k2 |α|≤s |α|≤s X Z Z X = ξαφˆ(ξ) 2dξ = φˆ(ξ) 2( ξα 2) dξ. n | | n | | | | |α|≤s R R |α|≤s Since both P ξα 2 and (1 + ξ 2)s are polynomials on ξ of the same degree, then |α|≤s | | | | there exist two positive constants C1,C2 for which Z 2 ˆ 2 2 s 2 C1 φ Hs φ(ξ) (1 + ξ ) dξ C2 φ Hs . k k ≤ Rn | | | | ≤ k k R 2 2 s 1/2 Which means that for s = 0, 1, 2,... the norms H and n ( )ˆ(ξ) (1+ ξ ) dξ k·k s R | · | | | are equivalent. Since the second norm is meaningful for any s R, we define the ∈ Sobolev s-norm of a function φ C∞(Ω) as ∈ 0 1 Z 2 ˆ 2 2 s φ Hs := φ(ξ) (1 + ξ ) k k Rn | | | | where φ is defined to vanish on Ωc (φ(x) = 0 if x / Ω). The completion of C∞(Ω) with ∈ 0 respect to the Sobolev s-norm is called the Sobolev s-space and is denoted by Hs(Ω). Note that for 0 < t < s one has continuous inclusions 2 Hs(Ω) Ht(Ω) H0(Ω) = L (Ω). ⊂ ⊂ It is clear that Ck(Ω) Hk(Ω). This property has a partial converse: ⊂ n Theorem 31 (Sobolev embedding). If φ Hk(Ω), then for all s < k we have ∈ − 2 f Cs(Ω). ∈ n α Proof. Fix k and s such that k > s + . We first show that the map D : Hk(Ω) 2 → H (Ω) is continuous as long as α s. Indeed, for φ Hk(Ω), k−|α| | | ≤ ∈ Z α 2 α 2 2 k−|α| D φ Hk−|α| = (D φ)ˆ (1 + ξ ) dξ k k Rn | | | | Z = ξαφˆ(ξ) 2(1 + ξ 2)k−|α|dξ. Rn | | | | 6.1 Sobolev Spaces 63 Since both (1 + ξ 2)k and ξα 2(1 + ξ 2)k−|α| are polynomials of degree 2k there exists | | | | | | C > 0 making α D φ H C φ H . k k k−|α| ≤ k k k We now want to show that if Dαφ H (Ω) then Dαφ C0(Ω) as long as α s. ∈ k−|α| ∈ | | ≤ If we prove this we would then have that φ Cs(Ω) as desired. Let x Ω, ∈ ∈ Z α ix·ξ α D φ(x) = e (D φ)ˆ(ξ)dξ | | Rn Z ix·ξ 2 −(k−|α|)/2 2 (k−|α|)/2 α = e (1 + ξ ) (1 + ξ ) (D φ)ˆ(ξ)dξ Rn | | | | Z 1/2 Z 1/2 (1 + ξ 2)−(k−|α|)dξ (Dαφ)ˆ(ξ) 2(1 + ξ 2)k−|α|dξ ≤ Rn | | Rn | | | | and since k α > n , there exists some C > 0 for which − k | 2 α α D φ(x) C D φ H . (6.1) | | ≤ k k k−|α| α ∞ Since D φ Hk−|α|(Ω), there exists a sequence ψj j C0 (Ω) such that ψj α ∈ { } α⊂ α → D φ in H (Ω). Equation (6.1) implies that ψj D φ uniformly and so D φ is k−|α| → continuous. Theorem 32 (Compactness). If s < t, then the inclusion Ht(Ω) Hs(Ω) is compact. ⊂ Proof. Saying that the inclusion is compact means that any bounded sequence φj j { } ⊂ Ht(Ω) has a subsequence φj k which is convergent in Hs(Ω). So let us start fixing { k } the sequence φj j Ht(Ω). The argument is as follows. We first show that if φj j { } ⊂ ˆ ˆ { } is bounded, φj Ht 1, then φj j and ∂ξi φj j are uniformly bounded on compact kn k ≤ { } { } subsets of R . By Arzela-Ascoli’s Theorem this implies that there is a subsequence ˆ n φjk k that converges uniformly on compact subsets of R . Using this, we then prove { } that φj k is a Cauchy sequence in Hs(Ω). Since by definition Hs(Ω) is complete, we { k } get that φj k is convergent in Hs(Ω). { k } n ∞ n Let ξ R and let χ C0 (R ) be so that χ(x) = 1 for all x Ω. ∈ ∈ ∈ φˆj(ξ) = (χ φj)ˆ(ξ) | | | · | = χˆ φˆj(ξ) |Z ∗ | = χˆ(ξ η)φˆj(η)dη Rn − Z 1/2 Z 1/2 2 2 −t 2 2 t χˆ(ξ η) (1 + η ) dη φˆj(η) (1 + η ) dη ≤ Rn | − | | | Rn | | | | = f(ξ) φj H k k t n ˆ where f(ξ) is a continuous function on R . Since φj Ht 1 this shows that φj j is n k k ≤ ˆ{ } uniformly bounded on compact subsets of R . Same argument works for ∂ξi φj j. Let { } 64 Heat Operator φj k be the subsequence given by Arzela-Ascoli’s Theorem. It only remains to show { k } that φj k is a Cauchy sequence in Hs(Ω). Fix ε > 0. For r > 0, { k } Z 2 ˆ ˆ 2 2 s φjk φj` Hs = φjk (ξ) φj` (ξ) (1 + ξ ) dξ k − k Rn | − | | | Z Z ˆ ˆ 2 2 s ˆ ˆ 2 2 s = φjk (ξ) φj` (ξ) (1 + ξ ) dξ + φjk (ξ) φj` (ξ) (1 + ξ ) dξ |ξ|≤r | − | | | |ξ|>r | − | | | (6.2) The idea is to pick r sufficiently large so that the second term on the RHS of (6.2) is n ˆ smaller than ε/2. Since ξ R : ξ r is compact and φjk k converges uniformly { ∈ | | ≤ } { } on compact subsets we can make the first therm on the RHS of (6.2) be smaller than ε/2. To choose r simply note that if ξ > r then | | (1 + ξ 2)t = (1 + ξ 2)s(1 + ξ 2)t−s (1 + ξ 2)s(1 + r2)t−s. | | | | | | ≥ | | Therefore, Z ˆ ˆ 2 2 s φjk (ξ) φj` (ξ) (1 + ξ ) dξ |ξ|>r | − | | | Z 1 ˆ ˆ 2 2 t 2 t−s φjk (ξ) φj` (ξ) (1 + ξ ) dξ ≤ (1 + r ) |ξ|>r | − | | | 1 2 φˆj φˆj ≤ (1 + r2)t−s k k − ` kHt 4 , ≤ (1 + r2)t−s and so we may pick r as large as we want to make 4 ε/2. (1+r2)t−s ≤ Sobolev spaces on compact manifolds. We now extend the definition of the Sobolev n spaces to a compact Riemannian manifolds (M, g). Let ui : Ui R M be an { ⊂ → } atlas of coordinate maps on M with U i compact, and write ρi : M [0, 1] for a { → }∞ partition of unity associated to Ui i. We define Hs(M) as the completion of C (M) { } 0 with respect to the norm 1 ! 2 X 2 φ H := ρiφ ui . k k s k ◦ kHs i It is not difficult to show that this definition is invariant under changes of coordinate charts and partitions of unity. All the results in this section extend to Sobolev spaces on compact manifolds in a natural manner. 6.2 Heat propagator Throughout this section we assume that (M, g) is a compact Riemannian manifold without boundary. The heat operator L := ∆ + ∂t acts on functions in C(M × 6.2 Heat propagator 65 (0, + )) that are C2 on M and C1 on (0, ). The Heat equation is given by ∞ ∞ ( Lu(x, t) = F (x, t)(x, t) M (0, + ) ∈ × ∞ . u(x, 0) = f(x) x M ∈ The homogeneous Heat equation is ( Lu(x, t) = 0 (x, t) M (0, + ) ∈ × ∞ . u(x, 0) = f(x) x M ∈ Lemma 33. If u(x, t) solves the homogeneous heat equation, then the function t 7→ u( , t) 2 is decreasing with t. k · kL Proof. The proof reduces to show that the t-derivative of the map is negative: Z d 2 u( , t) L2 = 2 ∂tu(x, t) u(x, t)ωg(x) dtk · k M Z = 2 ∆gu(x, t) u(x, t)ωg(x) − M 2 = 2 gu( , t) − k∇ · k 0. ≤ Lemma 34. The solution to the inhomogenous problem is unique. Proof. Suppose that both u1 and u2 are solutions to the homogeneous problem. Then u = u1 u2 solves − ( Lu(x, t) = 0 (x, t) M (0, + ) ∈ × ∞ . u(x, 0) = 0 x M ∈ R 2 The proof follows from the fact that M u(x, t) dx is a decreasing function of t while u(x, 0) = 0 for x M. It follows that u(x, 0) = 0 for all x M. ∈ ∈ 2 Proposition 35 (Duhamel’s principle). Let u, v : M (0, + ) R be C on M and × ∞ → C1 on (0, + ). Then, for any t > 0 and α, β such that [α, β] (0, t), we have ∞ ⊂ Z u(y, t α)v(y, α) u(y, t β)v(y, β) ωg(y) = M − − − Z β Z = Lu(y, t s) v(y, s) u(y, t s) Lv(y, s) ωg(y)ds. α M − − − Proof. Lu(y, t s) v(y, s) u(y, t s) Lv(y, s) = − − − = ∆gu(y, t s)v(y, s) ∆g(y, t s)Lv(y, s) + ∂su(y, t s)v(y, s) u(y, t s)∂sv(y, s) − − − − − − = ∆gu(y, t s)v(y, s) ∆g(y, t s)Lv(y, s) ∂s(u(y, t s)v(y, s)). − − − − − The result follows from integrating first with respect to x (this makes the first two terms cancel out) and then with respect to t. 66 Heat Operator We say that a fundamental solution of the heat equation is a continuous function 2 1 p : M M (0, ) R which is C with respect to x, C with respect to t and such × × ∞ → that Lyp = 0 and lim p( , y, t) = δy. t→0 · Proposition 36. The fundamental solution to the heat equation on M is unique and symmetric in the two space variables. Proof. Let p1 and p2 be two fundamental solutions. Fix x, z M. In Duhamel’s ∈ principle set u(y, t) = p1(x, y, t) and v(y, t) = p2(z, y, t). Using that Lyp1 = Lyp2 = 0 we get Z p1(x, y, t α)p2(z, y, α) p1(x, y, t β)p2(z, y, β) ωg(y) = 0. M − − − Let α 0 and β t. We then get → → p1(x, z, t) = p2(z, x, t). In particular, choosing p1 = p2 we get that the fundamental solution is symmetric. Since the fundamental solution is symmetric we deduce that p1(x, z, t) = p2(z, x, t) = p2(x, z, t) and so the fundamental solution is unique. Proposition 37. Let f C(M) and F C(M (0, + )). Then ∈ ∈ × ∞ Z Z t Z u(x, t) = p(x, y, t)f(y) dωg(y) + p(x, y, s)F (y, t s)dωg(y) M 0 M − solves the problem ( Lu = F, u( , 0) = f. · Proof. Fix x M. Apply Duhamel’s principle to u and v(y, t) = p(x, y, t) and get ∈ Z Z β Z u(y, t α)p(x, y, α) u(y, t β)p(x, y, β) ωg(y) = F (y, t s) p(x, y, s)ds. M − − − α M − Let α 0 and β t. → → R Remark 38. We stop to observe that p(x, y, t) dωg(y) = 1 x M, t (0, ). ( M ∀ ∈ ∈ ∞ Lu = F, This is because u(x, t) 1 solves ≡ u( , 0) = f. · Remark 39. For any t, s > 0 we get Z p(x, z, t + s) = p(x, y, t)p(y, z, s)ωg(y). M ( Lu = 0 This is because for fixed y M we have u(z, s) = p(z, y, s+t) solves . ∈ u( , 0) = p( , y, t) · · Remark 40. All these results can be proved for compact manifolds with boundary using an adaptation of Duhamel’s principle. One should use that both in the Dirichlet and R ∞ Neumann boundary conditions setting one has φ∂νψ dσ = 0 for all φ, ψ C (M) ∂M ∈ satisfying the appropriate boundary conditions. 6.3 Basis of eigenfunctions on compact manifolds 67 6.3 Basis of eigenfunctions on compact manifolds Given t > 0, let us define the heat propagator e−t∆g : L2(M) L2(M) by → Z −t∆g e f(x) := p(x, y, t)f(y) ωg(y). M Lemma 41. 1. e−t∆g e−s∆g = e−(t+s)∆g ◦ 2. e−t∆g is self-adjoint and positive. 3. e−t∆g is a compact operator. Proof. R 1. Follows from the fact that p(x, z, t + s) = M p(x, y, t)p(y, z, s)ωg(y). 2. For f, g L2(M), ∈ Z Z −t∆g e f, g = p(x, y, t)f(y) ωg(y) g(x) ωg(x) h i M M Z Z = p(y, x, t)f(y)g(x) ωg(x) ωg(y) M M Z Z = p(y, x, t)g(x) ωg(x) f(y) ωg(y) M M = f, e−t∆g (g) . h i This shows that e−t∆g is self-adjoint. To show that it’s positive it suffices to notice −t∆ − t ∆ 2 that e g f, f = e 2 g f 0. h i k k ≥ −t∆g 2 3. The operator e : L (M) H1(M) is continuous and the inclusion H1(M) → ⊂ L2(M) is compact. Their composition e−t∆g : L2(M) L2(M) is therefore −t∆g 2 → compact. To see that e : L (M) H1(M) is continuous simply note that 2 →2 −t∆g if fj j L (M) converges to 0 in L then e fj L2 j 0 and similarly { −}t∆⊂g k k → ∂x e fj 2 j 0. The last two statements can be shown by splitting M in k k kL → coordinate charts and pasting them by partition of unity. Lemma 42. As t 0 we have e−t∆g f f in L2(M). → → Proof. Z 2 Z Z 2 f p(x, y, t)f(y)ωg(y) = f(x) p(x, y, t)f(y)dωg(y) ωg(x) − M L2 M − M Z Z 2 = p(x, y, t)(f(x) f(y)) ωg(y) ωg(x) M M − Z Z 2 p(x, y, t) f(x) f(y) ωg(y)ωg(x). ≤ M M | − | 68 Heat Operator Since p(x, y, t) is uniformly bounded and p(x, y, t) f(x) f(y) 2 0 as t 0, we | − | → → conclude our result from the Dominated convergence theorem. Theorem 43. For any f L2(M) the function e−∆g f converges uniformly as t to ∈ → ∞ a harmonic function. In particular, it converges to a constant function when ∂M = . ∅ Proof. The strategy is to show first that e−t∆g f converges in L2(M). Then show that e−t∆g f converges uniformly to a continuos function f˜. Lastly, prove that e−t∆g f˜ = f˜. −t∆g If we do this, then 0 = L(e f˜) = Lf˜ = ∆gf˜. This would show that f˜ is harmonic and since M is compact we must have that f˜ is the constant function. −t∆g We know that e f 2 is a decreasing function of t, so it must converge to something. k kL Now, −t∆ −s∆ 2 −t∆ 2 −s∆ 2 − t+s ∆ 2 e g f e g f = e g f + e g f 2 e 2 g f . k − k k k k k − k k This shows that e−t∆g f e−s∆g f 0 as s, t . k − k → → ∞ Let x M. ∈ (e−(T +t)∆g f e−(T +s)∆g f)(x) 2 = e−T ∆g (e−t∆g f e−s∆g f)(x) 2 | − | | − | Z 2 −t∆g −s∆g = p(x, y, T )(e f e f)(y)ωg(y) M − −t∆g −s∆g 2 constant e f e f 2 . ≤ · k − kL We now show that e−t∆g f˜ = f˜: Z 2 −(t+s)∆g −t∆g 2 −s∆g (e f e f˜) (x) = p(x, y, t)(e f f˜)(y)ωg(y) | − | M − −s∆g 2 constant e f f˜ 2 . ≤ · k − kL Theorem 44 (Sturm-Liouville decomposition). For M compact, there exists a complete 2 orthonormal basis ϕ0, ϕ1,... of L (M) consisting of eigenfunctions of ∆g with ϕj { } having eigenvalue λj satisfying λ0 λ1 λ2 ... . ≤ ≤ ≤ → ∞ ∞ For every j we have ϕj C (M) and ∈ ∞ X −λj t p(x, y, t) = e ϕj(x)ϕj(y). j=0 Proof. We recall that so far we haven’t proved the existence of the fundamental solution p. Once we do so we will show that p is C∞ in the spatial variables. The proof that follows will yield that the smoothness of ϕj is that of p in the spatial variables and so ∞ it will follow that ϕj C (M). ∈ 6.3 Basis of eigenfunctions on compact manifolds 69 −t∆ −∆ t −k∆ −∆ k Our first claim is that e g = (e g ) for t > 0. If k Z, then e g = (e g ) . Fix 1 1 1 − ∆g q −∆ − ∆g −∆ ∈ p, q Z. Since (e q ) = e g , then e q = (e g ) q . We then need to generalize ∈ this for t R. Note that ∈ − p ∆ − p ∆ p p e−t∆g (e−∆g )t e−t∆g e q g + e q g (e−∆g ) q + (e−∆g ) q (e−∆g )t . k − k ≤ k − k |k −{z k} k − k =0 Observe that p p p (e−∆g ) q (e−∆g )t = sup β q βt 0 as t. k − k β∈spec(e−∆g ) | − | → q → On the other hand, p Z Z −t∆g − q ∆g 2 2 2 (e e )f L2 p(x, y, t) p(x, y, p/q) f(y) ωg(y)ωg(x) k − k ≤ M M | − | | | also converges to 0 by dominated convergence. −∆g Since e is a compact self-adjoint operator we know that it has eigenvalues β0 β1 ≥ ≥ βk 0 as n with respective eigenfunctions ϕ0, ϕ1,... forming a complete · · · ≥ → → ∞2 −t∆g −∆g t −t∆g t orthonormal basis of L (M). Since e = (e ) , we must have e ϕ0 = β ϕ0. R 0 From the fact that u(x, t) = M p(x, y, t)ϕ0(y)ωg(y) is a solution to the homogeneous equation ( Lu = 0 u( , 0) = ϕ0 · R 2 and that u(x, t) ωg(x) decreases with t we have β0 1. M ≤ Set λk := ln βk. Then − −t∆g −tλk e ϕk = e ϕk. −t∆g Since e ϕk is as solution of the heat equation for all k we get −t∆g −λkt 0 = L(e ϕk) = e (∆ϕk λkϕk), − which implies that ϕk is an eigenfunction of the Laplacian with eigenvalue λk. P∞ Since p(x, y, t) = p(x, , t), ϕk ϕk(y), and k=0h · i −t∆g −λkt p(x, , t), ϕk = e ϕk(x) = e ϕk(x) h · i we finally obtain that ∞ X −λkt p(x, y, t) = e ϕk(x)ϕk(y). k=0 R 2 Since M ϕkωg = 1 for all k we conclude the following Corollary 45. For every t > 0 Z ∞ X −λkt p(x, x, t)ωg(x) = e . M k=0 70 Heat Operator 6.4 Fundamental solution in Rn Eventhough we have been working on compact manifolds throughout this chapter, we digress briefly to inspire the form of the fundamental solution on compact manifolds. n Suppose p(x, y, t) is a fundamental solution for Heat equation in R . That is, p : n n 2 n 1 R R (0, + ) is continuous, C on R , C on (0, + ) × × ∞ ∞ ∆g n p + ∂tp = 0, R p(x, y, t) δx(y) t 0. → → R Then u(x, t) = n p(x, y, t)f(y)dy solves the homogeneous heat equation R ( (∆g n + ∂t)u = 0 R u( , 0) = f · n where f is a continuous and bounded function on R . Let us use the notation “ ˆ ” for the Fourier tranform with respect to the spatial variables: n n X X y 2uˆ(y, t) = y2uˆ(y, t) = ( D2 u)b(y, t) k k j xj j=1 j=1 n X 2 = ( ∂ u)b(y, t) = (∆g n u)b(y, t) − xj R j=1 = ∂ctu(y, t) = ∂tuˆ(y, t). − − n It follows that there exists a function h : R R such that → 2 uˆ(y, t) = h(y)e−tkyk so therefore h(y) =u ˆ(y, 0) = fˆ(y). We deduce that √ −tkyk2 −k 2tyk2/2 − n −k·k2/4tb uˆ(y, t) = fˆ(y)e = fˆ(y)e = fˆ(y) (2t) 2 e (y) · and so h − n −k·k2/4tib uˆ( , t) = f (2t) 2 e . · ∗ It follows that formally Z h − n −k·k2/4ti 1 −ky−xk2/4t u(y, t) = f (2t) 2 e (y) = n f(x)e dx. ∗ (4πt) 2 Rn We then should have 1 −ky−xk2/4t p(x, y, t) = n e . (4πt) 2 n 6.4 Fundamental solution in R 71 n n Proposition 46. The function p : R R (0, + ) [0, + ) × × ∞ → ∞ 1 −ky−xk2/4t p(x, y, t) = n e (4πt) 2 n is a fundamental solution for the heat equation on R . Proof. It is easy to check that ∆g n p + ∂tp = 0. Let us prove that p(x, y, t) R R → δx(y) t 0. We first need to show that n p(x, y, t)dy = 1 for all (x, t) n → R ∈ R (0, + ). Indeed, × ∞ Z Z ∞ Z p(x, y, t)dy = p(x, x + rξ, t)rn−1dξdr Rn 0 Sn−1(x) Z ∞ Z 2 1 − r n−1 = n e 4t r dξdr 0 Sn−1(x) (4πt) 2 Z ∞ Z 1 −s2 n−1 n−1 1 = n e s (4t) 2 (4t) 2 dξdr 0 Sn−1(x) (4πt) 2 Z ∞ Z 1 −s2 n−1 = n e s dξds π 2 0 Sn−1(x) Z 1 −kyk2 = n e dy π 2 Rn = 1. n Let f : R R be a bounded and continuous function. → Z f(x) p(x, y, t)f(y)dy = − Rn Z = p(x, y, t)(f(x) f(y))dy Rn − Z Z p(x, y, t) f(x) f(y) dy + p(x, y, t)(f(x) f(y))dy ≤ √ | − | n √ − B2 tR(x) R \B2 tR(x) Z ∞ Z n−1 sup f(x) f(y) + √ p(x, x + rξ, t) f(x) f(x + rξ) r dξdr. ≤ √ | − | n−1 | − | y∈B2 tR(x) 2 tR S (x) We now need to choose R sufficiently large so that the second term is as small as we wish. Once R is fixed the first term can be chosen to be small since t 0. → Z ∞ Z n−1 √ p(x, x + rξ, t) f(x) f(x + rξ) r dξdr = 2 tR Sn−1(x) | − | Z ∞ Z n−1 p(x, x + 2√ts, t)2 f ∞(2√ts) 2√t dξds ≤ R Sn−1(x) k k Z ∞ Z −s2 n−1 = e s 2 f ∞dξds R Sn−1(x) k k Z ∞ n−1 −s2 n−1 = 2vol(S ) f ∞ e s ds. k k R 72 Heat Operator 6.5 Existence of the fundamental solution on compact man- ifolds Let (M, g) be a compact manifold and choose ε > 0 such that the exponential map exp is a well defined diffeomorphism on Bε(0) TxM. We then identify a point x ⊂ y Bε(x) := expx(Bε(0)) with its polar coordinates (r, ξ), y = expx(rξ) where r (0, ) ∈ n−1 ∈ and ξ S (0) TxM. By performing a computation very similar to the one we ∈ ⊂ n carried when we expressed the Laplacian on R in terms of that of a radial operator and the Laplacian of Sn−1, once can show that using the geodesic polar coordinates y = (r, ξ) one gets 2 ∂r(√det g∂r) ∆g = ∂r + ∆g n−1 , − − √det g Sr (x) n−1 where gSn−1(x) is the metric induced on the geodesic sphere Sr (x) M of radius n ⊂ r (0, ε). Remember that the fundamental solution on R has the form (x, y, t) ∈ 2 7→ (4πt)−n/2ekx−yk /4t. Inspired on this formula we wish to find the fundamental solution on M, but in order to do this we need to work with the Riemannian distance function on M. For y = (r, ξ) Bε(x) we set dg(x, y) := r which is the length of the radial ∈ geodesic joining x and y. Consider the set ε := (x, y) M : y Bε(x), dg(x, y) < V { ∈ ∈ } and the function G : ε (0, + ) R V × ∞ → 1 2 G(x, y, t) := edg(x−y) /4t. (4πt)n/2 Unfortunately, not only G is not defined on all M but also one may check that (∆g + ∂t)G = 0. First, note that in geodesic polar coordinates y = (r, ξ) at x M one has 6 ∈ 1 2 G(x, y, t) := er /4t (4πt)n/2 and so (writing ∆g for the Laplacian on the spatial variable y) 2 ∂r(√det g∂rG) 2 ∂r√det g ∆gG = ∂ G = ∂ G ∂rG . − r − √det g − r − √det g Defining the function D : Bε(x) R → p det g(y) D(y) := rn−1 where g(y) is thought of as a matrix in the y = (r, ξ) geodesic polar coordinates at x we get 2 ∂rD n + 1 ∆gG = ∂ G ∂rG + − r − D r 2 n r r ∂rD = G + G. 2t − 4t2 2t D On the other hand, n r2 ∂tG = G. − 2t − 4t2 6.5 Existence of the fundamental solution on compact manifolds 73 This clearly shows that (∆g + ∂t)G = 0. 6 We then try to modify G and we do so by considering for each k N the function ∈ Sk : ε (0, + ) R V × ∞ → k Sk(x, y, t) := G(x, y, t) u0(x, y) + tu1(x, y) + + t uk(x, y) ··· and we hope we can choose the functions uj ε that make (∆g + ∂t)Sk = 0. Spoilers ∈ V alert: we won’t be able to get the 0, but we will be able to get an expression that decays k−n/2 to zero like t as t 0. Let us compute (∆g + ∂t)Sk: → k k k ∆gSk = ∆gG(u0 + + t uk) + G(∆gu0 + + t ∆guk) 2 gG, g(u0 + + t uk) g. ··· ··· − h∇ ∇ ··· i Note that since G is not a function of the angular variables ξ, then gG involves no ∇ k ∂ξj terms. Since ∂r, ∂ξj g = 0 and ∂r, ∂r g = 1 we get gG, g(u0 + + t uk) g = h k i r h i k h∇ ∇ ··· i ∂rG(∂ru0 + + t ∂ruk) = (∂ru0 + + t ∂ruk)G. It then follows that ··· − 2t ··· k−1 r ∂rD k (∆g + ∂t)Sk = G u1 + . . . kt uk + (u0 + + t uk) (6.3) · 2t D ··· r k k + (∂ru0 + + t ∂ruk) + ∆gu0 + + t ∆guk . (6.4) t ··· ··· Although we are not able to build the functions uj so that (∆g + ∂t)Sk = 0 we will be able to build them so that k (∆g + ∂t)Sk = G t ∆guk (6.5) · · k−n/2 which makes (∆g + ∂t)Sk vanish to order t . Rearranging the terms in (6.3) one gets the following system of equations that ensure (6.5). We get r ∂rD r∂ru0 + u0 = 0, (6.6) 2 D r ∂rD r∂ruj + + j uj + ∆g uj−1 = 0, j = 1, . . . , k. (6.7) 2 D −1/2 Equation (6.6) gives u0 = fD where f is a function of the angular variables ξ. We wish that u0(x, y) be defined at x = y, or equivalently r = 0. This means that f must be constant and we set it to be 1. Therefore, −1/2 u0(x, y) = D (y). Since we chose f 1 we get ≡ u0(x, x) = 1. 0 We now need to prove the existence of the other ujs. Instead of solving (6.7) we note −j −1/2 that vj = f r D with f = f(ξ) solves the simpler problem r ∂rD r∂rvj + + j vj = 0. 2 D 74 Heat Operator One can then check that −j −1/2 uj := h r D with h = h(r) solves (6.7) as long as 1/2 j−1 ∂rh = D ∆guj−1 r . − Let γ be the geodesic on M joining x and y. Since ∆guj−1( , y) is a functions of r along · γ, we may find h by integration in r Z r 1/2 j−1 h(r) = D (γ(s))∆guj−1(γ(s), y) s ds. 0 Finally, Z r −j −1/2 1/2 j−1 uj(x, y) = (d(x, y)) D (y) D (γ(s))∆guj−1(γ(s), y) s ds. − 0 ∞ It is easy to see that by induction uj C ( ). ∈ V Let us record what we proved so far: ∞ Proposition 47. The family of functions uk C ( ) defined by the recursion formu- las ∈ V −1/2 u0(x, y) = D (y) Z r −k −1/2 1/2 k−1 uk(x, y) = (d(x, y)) D (y) D (γ(s))∆guj−1(γ(s), y) s ds − 0 satisfies k LySk = G t ∆guk for all k = 0, 1,... · · 1 where Ly = ∆g,y + ∂t. In particular, u0(x, x) = 1 and u1(x, x) = 6 Rg(x). We now wish to extend the definition of Sk to all of M. We then introduce a bump function α C∞(M M, [0, 1]) with α 0 on c and α 1 on . We may now ∈ × ≡ Vε ≡ Vε/2 define Hk : M M (0, + ) R × × ∞ → Hk(x, y, t) := α(x, y)Sk(x, y, t). ∞ Lemma 48. For k > m/2 the function Hk C (M M (0, + )) satisfies ∈ × × ∞ ` a) LyHk C (M M [0, + )) for 0 ` < k n/2. ∈ × × ∞ ≤ − b) For every x M,Hk(x, y, t) δx(y) as t 0 for all y M. ∈ → → ∈ Proof. a) We prove this statement for ` = 0. The problem with establishing the continuity of Hk is to show we have it at t = 0. By the definition of the bump function we have c c Hk 0 on (0, + ). So we may extend the definition of Hk to [0, + ) to be ≡ Vε × ∞ Vε × ∞ 0 at t = 0. Since α 1 on and ≡ Vε/2 1 −r2/4t k LyHk = e t ∆guk 0 as t 0, (4πt)n/2 → → 6.5 Existence of the fundamental solution on compact manifolds 75 we may also extend Hk on ε/2 [0, + ) by setting it to be 0 at t = 0. It only remains V c× ∞ to deal with the domain [0, + ). We then have V ∩ Vε/2 × ∞ LyHk = α (∆g + ∂t)Sk + ∆gα Sk 2 gα, gSk g · · − h∇ ∇ i 1 2 = e−r /4tβ(x, y, t) (4πt)n/2 ∞ 1 where β(x, y, t) C (M M (0, + )) has at most a pole of order t at t = 0 coming ∈ × c × ∞ from gSk. Since on [0, + ) we have r > ε/2, we may also extend Hk to ∇ V ∩ Vε/2 × ∞ be 0 at t = 0. k b) Let f C(M). Since Hk = α G (u0 + . . . t uk), we are interested in understanding ∈ the behavior of the integral Z α(x, y)G(x, y, t)uj(x, y)f(y) dvg(y) M Z Z = G(x, y, t)uj(x, y)f(y) dvg(y) + α(x, y)G(x, y, t)uj(x, y)f(y) dvg(y) c Bε/2(x) Bε/2(x) 1 −r2/4t c First note that since G(x, y, t) = (4πt)n/2 e and r > ε/2 on Bε/2(x), we have that as t 0 → Z α(x, y)G(x, y, t)uj(x, y)f(y) dvg(y) 0. c → Bε/2(x) We now deal with the other term Z α(x, y)G(x, y, t)uj(x, y)f(y) dvg(y) = Bε/2(x) Z 1 −d2(x,y)/4t = n/2 e uj(x, y)f(y)dvg(y) Bε/2(x) (4πt) Z 1 −kvk2/4t = n/2 e uj(x, expx(v))f(expx(v))J(v)dv TxM (4πt) to get the last equality we passed to normal coordinates y v with y = expx(v), we → c wrote J is the jacobian for the change of variables, and we set uj 0 on Bε/2(0) . Since 1 −kvk2/4t ≡ n (4πt)n/2 e = p(0, v, t) is the fundamental solution to the heat equation on R , we get that as t 0 → Z α(x, y)G(x, y, t)uj(x, y)f(y) dvg(y) uj(x, x)f(x). Bε/2(x) → Using that u0(x, x) = 1 we get as t 0 → Z α(x, y)G(x, y, t)u0(x, y)f(y) dvg(y) f(x). Bε/2(x) → For any other values of j we get as t 0 → Z j α(x, y)G(x, y, t)t uj(x, y)f(y) dvg(y) 0. Bε/2(x) → 76 Heat Operator Before we find the actual expression for the fundamental solution it is convenient to understand how the Heat operator acts on convolutions. Let us define the convolution operation for functions F,H C0(M M [0, )) by ∈ × × ∞ Z t Z F H (x, y, t) := F (x, z, s)H(z, y, t s)ωg(z)ds. ∗ 0 M − We first note that if F C0(M M [0, )), then ∈ × × ∞ ` F Hk C (M M (0, )) for ` < k n/2. ∗ ∈ × × ∞ − This is easy to check, the only problem being the discontinuity of Hk at t = 0. We now explore how the heat operator Ly = ∆g + ∂t acts on F Hk(x, y, t). First observe that ,y ∗ ∂t(F Hk)(x, y, t) ∗ Z t Z = ∂t F (x, z, s)Hk(z, y, t s)ωg(z)ds 0 M − Z Z t Z = lim F (x, z, s)Hk(z, y, t s)ωg(z) + F (x, z, s)∂tHk(z, y, t s)ωg(z)ds s→t M − 0 M − Z t Z = F (x, y, t) + F (x, z, s)∂tHk(z, y, t s)ωg(z)ds, 0 M − and so Z t Z Ly(F Hk)(x, y, t) = = F (x, y, t) + F (x, z, s) LyHk(z, y, t s)ωg(z)ds ∗ 0 M − = F (x, y, t) + F (LyHk)(x, y, t). ∗ We then look for fundamental solutions of the form p = Hk F Hk − ∗ for some suitable choice of F . Note that Lyp = Ly(Hk F Hk) − ∗ = LyHk Ly(F Hk) − ∗ = LyHk F F (LyHk). (6.8) − − ∗ This suggest that we consider, at least formally, ∞ X j+1 ∗j Fk := ( 1) (LyHk) . − j=1 Indeed, if we had that this series is convergent then from (6.8), for p = Hk Fk Hk, − ∗ ∞ ∞ X j+1 ∗j X j+1 ∗j Lyp = LyHk ( 1) (LyHk) ( 1) (LyHk) (LyHk) = 0. (6.9) − − − − ∗ j=1 j=1 6.5 Existence of the fundamental solution on compact manifolds 77 ` Lemma 49. For all ` < k n/2 the series Fk C (M M [0, + )). In addition, − ∈ × × ∞ for t0 > 0 there exists a constant C = C(t0) such that k−n/2 Fk( , , t) ∞ Ct for all t [0, t0]. k · · kL (M×M) ≤ ∈ Proof. As we did in the proof of Lemma 48, using that gα and ∆gα have their supper ∇ away from the diagonal in M M, it is easy to get the existence of a constant C = A(t0) × such that k−n/2 LyHk( , , t) ∞ Ct for all t [0, t0]. k · · kL (M×M) ≤ ∈ In particular, k−n/2 LyHk ∞ Ct k kL (M×M×[0,t0]) ≤ 0 We claim that for j = 1, 2,... and t [0, t0] ∈ k−n/2 j−1 j−1 ∗j C (Ct0 ) volg(M) k− n +j−1 (LyHk) ( , , t) L∞(M×M) n n t 2 . k · · k ≤ (k n/2 + j 1) ... (k 2 + 2)(k 2 + 1) − − − − (6.10) Indeed, by induction, assuming that it is true for j 1 we get for x, y M − ∈ ∗j (LyHk) (x, y, t) | | ≤ Z t Z ∗(j−1) (LyHk) (x, z, s) LyHk(z, y, t s) ωg(z)ds ≤ 0 M | || − | Z t Z k−n/2 j−2 j−2 C (Ct0 ) volg(M) k− n +j−2 k−n/2 t 2 Ct ωg(z)ds ≤ (k n/2 + j 2) ... (k n + 2)(k n + 1) 0 0 M − − − 2 − 2 k−n/2 j−1 j−2 Z t C (Ct0 ) volg(M) k− n +j−2 = volg(M) t 2 ds. (k n/2 + j 2) ... (k n + 2)(k n + 1) − − − 2 − 2 0 P ∗j Using (6.10) the ratio test shows that j (LyHk) L∞ is convergent if k > n/2, and P j+1 ∗j k k so in particular ( 1) (LyHk) converges to a continuous function. j − The same kind of argument can be carried for the derivatives of LyHk. We have showed that Fk is well defined. We then prove Proposition 50. The function p = Hk Fk Hk is a fundamental solution for the Heat − ∗ equation for all k > n/2 + 2 . 2 Proof. For k > n/2 + 2 we know that p C (M M (0, + )). Since Fk := P∞ j+1 ∗j ∈ × × ∞ ( 1) (LyHk) , j=1 − Lyp = Ly(Hk F Hk) = LyHk Ly(F Hk) = LyHk F F (LyHk) = 0. − ∗ − ∗ − − ∗ Let us now prove that p(x, y, t) δx(y) as t 0: → → Z Z lim p(x, y, t)f(y) ωg(y) = lim Hk(x, y, t) Fk Hk(x, y, t) f(y) ωg(y) t→0 M t→0 M − ∗ Z = f(x) lim Fk Hk(x, y, t) f(y) ωg(y) − t→0 M ∗ 78 Heat Operator k−n/2 By Lemma 49 we have Fk( , , t) L∞(M×M) Ct for all t [0, t0] and so the kernel −(k−n/2) k · · k ≤ ∈ Rk := t Fk is uniformly bounded for (x, y, t) M M [0, t0]. Therefore, since ∈ × × Hk(x, y, t) δx(y) as t 0 we get that for k > n/2 → → Z Z k−n/2 lim Fk Hk(x, y, t) f(y) ωg(y) = lim t Rk Hk(x, y, t) f(y) ωg(y) = 0. t→0 M ∗ t→0 M ∗ Theorem 51. There exists > 0 such that the fundamental solution for the heat equa- tion has expansion 2 k −dg(x,y)/4t e X j k+1 p(x, y, t) = n/2 t uj(x, y) + O(t ) , (4πt) j=0 1 for all x, y M with dg(x, y) ε/2. In addition, u0(x, x) = 1 and u1(x, x) = Rg(x). ∈ ≤ 6 Proof. Since p = Hk Fk Hk is a fundamental solution for the heat equation for any − ∗ k > n/2 + 2 and such solution is unique, we have that the definition of p is independent k−n/2 of k. Note also that Hk Fk Hk C (M M (0, + )) for all k > n/2 which − ∗ ∈ × × ∞ shows that p C∞(M M (0, + )). ∈ × × ∞ k−n/2 1 Since Fk( , , t) L∞(M×M) Ct and Hk L (M M (0, t)), we get Fk k · · k k+1≤−n/2 ∈ × × k ∗ Hk( , , t) L∞(M×M) Dt for all t t0 for some given small t0. The asymptotics · · k ≤ 2 ≤ e−dg(x,y)/4t k follow from the definition of Hk = α(x, y) (u0 + + t uk). (4πt)n/2 · ··· 6.6 Fundamental solution in Riemannian coverings ˜ Let (M, gM˜ ) be a Riemannian manifold and let Γ be a discrete group acting properly and freely on M˜ . Write (M, gM ) for the compact quotient manifold M = M/˜ Γ. Since M is compact, there exists a relatively compact open set D M˜ such that M˜ = γ∈Γ γD. ⊂ ∪ Also, the group Γ is finitely generated and so Γ is a countable set. The construction of the fundamental solution for the heat equation on a compact mani- fold can also be carried on M˜ with almost no modification. Since M˜ = γ∈Γ γD with D ∪ relatively compact we can choose ε > 0 so that d(x, y) < ε implies that y lies in a normal coordinate neighborhood of x. On the sets ε := (x, y) M˜ M˜ : dg (x, y) < ε and V { ∈ × M˜ } defines the parametrix Hk as in the previous section. Then p˜ C∞(M˜ M˜ (0, + )) ∈ × × ∞ ∞ X j+1 ∗j p˜ := Hk ( 1) (LyHk) Hk − − ∗ j=1 is a fundamental solution for all k > n/2 + 2. We need a few results before we explain the relation between the heat kernel on M˜ and that of M. First, by comparison with spaces of contant curvature one has the following upper bound for the volume of geodesic balls (which we won’t prove): 6.6 Fundamental solution in Riemannian coverings 79 ˜ Lemma 52. Let (M, gM˜ ) be a Riemannian manifold whose all sectional curvatures are bounded below by κ. Then, for each x M˜ and r inj (M˜ ), one has ∈ ≤ C2 r volg (Br(x)) C1e M˜ ≤ where C1,C2 > 0 depend on κ and dimM˜ . Lemma 53. Set ND := # γ Γ: D γD = . For each x, y D and r > diam(D) { ∈ ∩ 6 ∅} ∈ volg B2r(x) # γ Γ: B (x) γy = N M˜ . r D vol { ∈ ∩ 6 ∅} ≤ gM˜ (D) Proof. For x, y D suppose that there is a number ` of Γ-translates of y inside Br(x). ∈ Since r > diam D, then B2r(x) contains ` translates of D. However, any point of M˜ is contained in at most ND translates of D. In consequence, ` volg (D) · M˜ volgM˜ (B2r(x)). ND ≤ Writep ˜ : M˜ M˜ (0, + ) R for the fundamental solution of the heat equation × × ∞ → in (M,˜ g ˜ ) and p : M M (0, + ) R for the fundamental solution of the heat M × × ∞ → equation in (M, gM ), where we continue to write M = M/˜ Γ. We also write π : M˜ M → for the covering map. Proposition 54. The fundamental solution for the heat equation on (M, g) is given by p : M M (0, + ) R × × ∞ → X p(x, y, t) = p˜(˜x, γ y,˜ t) γ∈Γ where x,˜ y˜ are such that π(˜x) = x, π(˜y) = y. The sum on the right hand side converges uniformly on D D [t0, t1] with 0 < t0 t1. × × ≤ Proof. We first show that the series converges: 2 X −n/2 X −dg (˜x,γy˜)/4t p˜(˜x, γ y,˜ t) Ct e M˜ ≤ γ∈Γ γ∈Γ ∞ 2 −n/2 X X −dg (˜x,γy˜)/4t Ct e M˜ ≤ j=1 {γ∈Γ: γy˜∈Bjr(˜x)\B(j−1)r(x)} ∞ −n/2 X ND −(j−1)2r2/4t Ct volgM˜ (B2jr(x))e ≤ volg (D) j=1 M˜ ∞ 2 2 −n/2 X ND 2C2jr −(j−1) r /4t CC1t e e ≤ volg (D) j=1 M˜ where to get the second inequality we decomposed M into rings whose boundary are geodesic spheres centred a t x of radious jr with r > diam(D) and j = 1, 2 ... . We 80 Heat Operator then used that for Bjr(x) B (x) one has dg (x, y) > (j 1)r, and we also used the \ (j−1)r M˜ − previous Lemma to bound the number of Γ-translates of y inside Bjr(x). We now show that p is a fundamental solution: Fix (x, t) M (0, + ) and consider the functionq ˜ : M˜ R defined byq ˜(˜y) := P ∈ × ∞ → γ∈Γ p˜(˜x, γ y,˜ t) wherex ˜ is so that πx˜ = x. It is clearly invariant under the deck transformation group Γ and so there exists a unique function q : M R making → q π =q ˜. Since q is unique then p(x, y, t) = q(y). We then have, fory ˜ such that ◦ πy˜ = y, ∗ ∆gM ,yq(y) = ∆gM q(πy˜) = π ∆gM q(˜y) ∗ = ∆gM˜ π q(˜y) = ∆gM˜ q˜(˜y) X = ∆gM˜ ,y˜p˜(˜x, γy,˜ t). γ∈Γ Since (∆g ,y + ∂t)˜p = 0, it follows that Lyq(y) = 0. To see that p(x, y, t) δx(y), M˜ → simply note that X p(x, y, t) δx˜(γy˜). → γ∈Γ It only remains to note that ( ( X 1 γ :x ˜ = γy˜ X 1 x = y δx˜(γy˜) = ∃ if and only if δx˜(γy˜) = , γ∈Γ 0 else γ∈Γ 0 else P and so γ∈Γ δx˜(γy˜) = δx(y). Flat Torus. From the expression for the fundamental solution of the heat equation n on R it follows that the fundamental solution for the heat equation on the flat torus n T = R /Γ is given by 1 X −ky˜+γ−x˜k2/4t pT(x, y, t) = n e 2 (4πt) γ∈Γ wherex, ˜ y˜ are such that π(˜x) = x, π(˜y) = y for the corresponding covering map n n π :(R , g n ) (R /Γ, g n /Γ). R → R Hyperbolic surfaces. Let (M, g) be a hyperbolic compact surface which is realized as the quotient H/Γ where H is the hyperbolic plane and Γ is a discrete subgroup acting properly and freely. To get the fundamental solution on M = H/Γ we just need to know that of H. We won’t prove the next result but you may find it in the book by Buser. Let d : H H R be the hyperbolic distance. H × → Proposition 55. The function p : H H (0, + ) R given by × × ∞ → Z ∞ −r2/4t √2 −t/4 re pH(x, y, t) = 3/2 e p dr (4πt) d (z,w) cosh r cosh d (z, w) H − H is a fundamental solution for the heat equation on H. 6.6 Fundamental solution in Riemannian coverings 81 We note that z w 2 cosh d (z, w) = 1 + | − | . H 2Imz Imw Negatively curved manifolds. If (M, g) is a compact manifold with negative sec- tional curvature everywhere, then by Hadamard’s Theorem for every x M the ex- ∈ n ponential map π := expx : TxM M is a covering map. Identifying TxM with R , → n n = dim M, we may think of (M, g) as the quotient of R by the deck transformation n ∗ group Γ associated to π. The metric we use on R is exactly π g. One may define the Dirichlet domain n DDir := y˜ R : dπ∗g(0, y˜) < dπ∗g(0, γy˜) γ Γ Id. { ∈ ∀ ∈ \ } We can add to DDir a subset of ∂DDir = DDir int(DDir) to obtain a fundamental n \ domain D which has the property that R is the disjoint union of the γD as γ ranges in Γ. One may then identify every point x M with a unique pointx ˜ D and ∈ ∈ X p(M,g)(x, y, t) = p(Rn,π∗g)(˜x, γ y,˜ t). γ∈Γ CHAPTER 7 Eigenvalues Throughout this section we assume that M is compact. Further, if M has a boundary, then we impose either Dirichlet or Neumann boundary conditions. Let ϕ0, ϕ1,... be an orthonormal basis of eigenfunctions of the Laplacian with respective eigenvalues λ0 λ1 ... , satisfying the corresponding boundary conditions when needed. ≤ ≤ 7.1 Self-adjoint extension of the Laplacian ∞ ∞ We know that the Laplacian ∆g : C (M) C (M) is formally self-adjoint. The → purpose of this section is to prove that the Laplacian admits a self-adjoint extension to 2 ∗ H (M) in the sense that the domains of ∆g and ∆g coincide. To show this, we first introduce a characterization of H1(M) and H2(M). Before we do this, we note that ∞ the Sobolev space H1(M) can be defined as the completion of C (M) with respect to the inner product u, v H := u, v g + gu, gv g h i 1 h i h∇ ∇ i ∞ ∞ for u, v C (M). Similarly, Hk(M) is the completion of C (M) with respect to ∈ k k u, v H := u, v H + u, v g. h i k h i k−1 h∇g ∇g i Proposition 56. ∞ n 2 X k 2 o Hk(M) = f L (M): λ f, ϕj < . ∈ j h i ∞ j=0 Proof. We give the proof of the characterization of H1(M). The one for Hk(M) is anal- 2 P∞ 2 o ogous. Define := f L (M): λj f, ϕj < . A { ∈ j=0 h i ∞ 84 Eigenvalues We first prove that H1(M). For f and k N consider the smooth approxima- A ⊂ ∈ A ∈ tion in L2(M) k k X S := ajϕj aj := f, ϕj . f h i j=0 Since Sk C∞(M), f ∈ k k l 2 k l k l X 2 (S S ) 2 = S S , ∆g(S S ) g = λja . k∇ f − f kL h f − f f − f i j j=l+1 k l 2 Pk 2 It follows that Sf Sf H1 = j=l+1(1 + λj)aj 0 as l, k . Since H1(M) is kk − k → → ∞ complete limk S = f H1(M) and so H1(M). f ∈ A ⊂ k k k k H1 P H1 Next we prove that is closed in H1(M), that is = . Let f = j ajϕj . ∞ A A A ∞ ∈ A Since C (M) is dense in H1(M), there exists a sequence fj j C (M) such that { } ⊂ fj f in H1(M). We may also consider the Fourier expansion for each fj. Say →P∞ (j) 2 fj = k=0 bk ϕk in L (M). Since fj f H1 j 0, we get fj f L2 j 0. Therefore P (j) k − k (j) → k − k → (b ak)ϕk 2 j 0, and this gives b j ak for all k. k k k − kL → k → In addition, (j) ∆gfj, ϕk g = fj, ∆gϕk g = λkb h i h i k P (j) and so ∆gfj = k λkbk ϕk. It follows that 2 2 2 fj = fj 2 + gfj 2 k kH1 k kL k∇ kL 2 = fj L2 + fj, ∆gfj g k∞ k h i X (j) 2 = (1 + λk)(bk ) k=0 ` X (j) 2 (1 + λk)(b ) ≥ k k=0 for all ` N. Since fj f H1 j 0, we know fj H1 j f H1 . In particular, since 2 ∈ P` k − (jk) 2 → k (kj) → k k fj H1 k=0(1 + λk)(bk ) for all ` N and bk j ak for all k N, we deduce k k2 ≥ P` 2 ∈ → P∈ ∞ 2 f H1 k=0(1 + λk)ak for all ` N. We then know that both k=0(1 + λk)ak k k P∞≥ 2 ∈ P∞ 2 and k=0 ak converge, which yields the convergence of k=0 λkak. We have proved k k H1 . A ⊂ A ⊥ Having shown that is closed in H1(M), all that remains to be proved is that = 0 A ⊥ ∞ A { } in H1(M). Let f H1(M) . Since C (M) is dense in H1(M) and is closed ∈ ∩ A ∞ ⊥ A in H1(M), there exists fj j C (M) such that fj f H j 0, and so in { } ⊂ ∩ A k − k 1 → particular, fj f 2 j 0. By Green’s identities k − kL → fj, ϕk H = fj, ϕk g + gfj, gϕk g = fj, ϕk g + fj, ∆gϕk g = (1 + λk) fj, ϕk g. h i 1 h i h∇ ∇ i h i h i h i Since fj, ϕk H = 0 for all k, it follows that f, ϕk g = 0 for all k and so fj = 0. This h i 1 h i shows that f 0. ≡ 7.1 Self-adjoint extension of the Laplacian 85 P ∞ P Note that if f = j ajϕj C (M), then ∆gf = j ajλjϕj and so ∆gf L2 = ∈ 2 k k f H (M). This suggests we define an extension ∆˜ g : H2(M) L (M) of the Laplacian. k k 2 P → For f H2(M), f = ajϕj, we set ∈ j ∞ X ∆˜ gf := ajλjϕj. j=0 Before we present the properties of the extension it is convenient to introduce a bilinear 1 ∞ ∞ form in H (M). Consider the bilinear form on Dg : C (M) C (M) R × → Dg(f, h) := gf, gh g. h∇ ∇ i ∞ Given f, h H1(M), there exist sequences fj j, hj j C (M) such that fj j f ∈ { } { } ⊂ → and hj j h in H1(M). We then define the bilinear form on H1(M) H1(M) by → × Dg(f, h) := lim Dg(fj, hj). j→∞ 2 Theorem 57. The extension ∆˜ g : H2(M) L (M) of the Laplacian has the following properties. → 1. ∆˜ gf, h g = Dg(f, h) for all f H2(M) and h H1(M). h i ∈ ∈ 2. ∆˜ g is self-adjoint. Proof. P P k Pk 1) Let f = j ajϕj H2(M) and h = j bjϕj H1(M). If we set Sf = j=0 ajϕj k Pk ∈ k k∈ and S = bjϕj, we get that S k and S k are Cauchy sequences in H1(M). h j=0 { f } { h} Then, ∞ k k k k k k X Dg(f, h) = lim D(Sf ,Sh) = lim gSf , gSh g = lim ∆gSf ,Sh g = λjajbj. k→∞ k→∞h∇ ∇ i k→∞h i j=0 In particular, by definition, ∞ ∞ ∞ X X X ∆˜ gf, h g = ajλjϕj, bjϕj = λjajbj = D(f, h). h i g j=0 j=0 j=0 2) From the previous part we know that ∆˜ g is formally self-adjoint. Indeed, if f, h ∈ H2(M), ∆˜ gf, h g = Dg(f, h) = Dg(h, f) = ∆˜ gh, f g. h i h i Set ∗ n 2 2 o Dom(∆˜ ) := u L (M): hu L (H) such that f, hu g = ∆˜ gf, u g f H2(M) . g ∈ ∃ ∈ h i h i ∀ ∈ Note that if such hu exists, then hu is unique. Indeed, if there were two such choices (1) (2) (1) (2) hu and hu then f, hu g = ∆˜ gf, u g = f, hu g for all f H2(M). Since H2(M) is 2 h i h i h i ∈ dense in L (M) we must have h1 = h2. This shows that we may define the operator ∆˜ ∗ : Dom(∆˜ ∗) L2(M) g g → 86 Eigenvalues ˜ ∗ ∆g u := hu. ˜ ˜ ∗ ˜ In order to show that ∆g is self-adjoint we need to prove that Dom(∆g) = Dom(∆g) where Dom(∆˜ g) = H2(M). ∗ Since ∆˜ g is formally self-adjoint we have Dom(∆˜ g) Dom(∆˜ ). Indeed, if u ⊂ g ∈ Dom(∆˜ g), then for all f H2(M) we have ∆˜ gf, u g = f, ∆˜ gu g and so trivially ∈ h i h i we may set hu := ∆gu. ˜ ∗ P Let us now prove the converse inclusion. Let u Dom(∆g), say u = j ajϕj. Then P 2 ∈ there exists a unique hu = bjϕj L (M) such that ∆˜ gf, u g = f, hu g for all j ∈ h i h i f H2(M). In particular, setting f = ϕj we get ∈ λjaj = λj ϕj, u g = ∆˜ gϕj, u g = ϕj, hu g = bj h i h i h i P 2 2 P 2 2 which gives λjaj = bj. Therefore λ a = b = hu 2 < and so by the j j j j j k kL ∞ characterization of H2(M) we got that u H2(M) = Dom(∆˜ g) as we wanted. ∈ 7.2 Characterization Let (M, g) be a compact Riemannian manifold and write λ1 λ2 ... for the eigenval- ≤ ≤ ues of the Laplacian repeated according to multiplicity (for any initial problem). Write 2 ϕ1, ϕ2,... for the corresponding L -normalized eigenfunctions. Since we require ϕi = 0 and ∆g(1) = 0.1 we have the following easy remark: 6 λ1 = 0 if ∂M = , ∅ λ1 > 0 Dirichlet boundary conditions, λ1 = 0 Neumann boundary conditions. ⊥ Theorem 58. For k N and Ek(g) := ϕ1, ϕs, . . . , ϕk−1 , ∈ { } nDg(φ, φ) o λk = inf : φ H1(M) Ek(g) . φ 2 ∈ ∩ k kg The infimum is achieved if and only if φ is an eigenfunction of eigenvalue λk. P∞ Proof. Fix φ H1(M) Ek(g) and assume it has expansion φ = ajϕj. Fix ` N, ∈ ∩ j=1 ∈ ` ` X X 0 Dg φ ajϕj , φ ajϕj ≤ − − j=1 j=1 ` ` X X = Dg(φ, φ) 2 ajDg(φ, ϕj) + ajaiDg(ϕj, ϕi) − j=1 i,j=1 ` ` X X = Dg(φ, φ) 2 aj φ, ∆gϕj + ajai ϕj, ∆gϕl g − h i h i j=1 i,j=1 ` X 2 = Dg(φ, φ) λja . − j j=1 7.2 Characterization 87 P` 2 Therefore, we get Dg(φ, φ) λjα for all ` N and so ≥ j=1 j ∈ ∞ ∞ ∞ X 2 X 2 X 2 2 Dg(φ, φ) λja λja λk a = λk φ (7.1) ≥ j ≥ j ≥ j k kg j=1 j=k j=k and so Dg(φ, φ) λk ≤ φ 2 k kg for all φ Ek(g). If φ is an eigenfunction of eigenvalue λk, then according to Corollary ∈ 18 Dg(φ, φ) φ, ∆gφ g = h i = λ φ 2 φ 2 k k kg k kg and so the infimum is achieved. On the other hand, if the infimum is achieved, from (7.1) follows that ∞ ∞ X 2 X 2 λjaj = λk aj j=k j=k and therefore ∞ X 2 (λj λk)a = 0. − j j=k If λj = λk then aj = 0 and so φ must be a linear combination of eigenfunctions with 6 eigenvalue λk. The previous theorem is a nice characterization of the eigenvalues, but in practise one need to know the eigenfunctions to use. Since finding the eigenfunctions is a much harder problem than determining the eigenvalues it is better to try to understand how to characterize the eigenvalues without using any eigenfunctions. We proceed to give a min-max characterization of the eigenvalues in terms of infimums and supremums over vectors spaces that are independent of the eigenfunctions. Theorem 59 (Max-Min Theorem). For k N let k−1 be the collection of all subspaces ∈ V V C∞(M) of dimension k 1. Then ⊂ − Dg(φ, φ) λk sup inf . ⊥ 2 ≥ V ∈V φ∈(V ∩H1(M))\{0} φ k−1 k kg Proof. Fix V k−1 and let ψ1, . . . , ψk−1 be a basis of orthonormal smooth func- ∈ V Pk D E tions of V . We claim that there exists φ˜ = aiϕi = 0 so that φ,˜ ψj = 0 for all i=1 6 j = 1, . . . , n k. Indeed, the existence of such function is equivalent to finding a1, . . . , ak Pk − so that ai ϕi, ψj = 0 for j = 1, . . . , k 1. This forms a system of k 1 equations i=1 h i − − with k unknowns, so it must have a solution. 88 Eigenvalues We then have Dg(φ, φ) Dg(φ,˜ φ˜) inf ⊥ 2 2 φ∈(V ∩H1(M))\{0} φ g ≤ φ˜ k k k kg k 1 X = aiaj gϕi, gϕj g φ˜ 2 h∇ ∇ i k kg ij=1 k 1 X = aiaj ϕi, ∆gϕj g φ˜ 2 h i k kg ij=1 k 1 X 2 = ai λi φ˜ 2 k kg i=1 λk. ≤ Theorem 60 (Min-Max Theorem). For k N let k be the collection of all subspaces ∈ V V C∞(M) of dimension k. Then ⊂ Dg(φ, φ) λk inf sup 2 . ≤ V ∈Vk φ∈V φ k kg Pk Proof. Note that V∆ := span ϕ1, . . . , ϕk k and for φ = ajϕj V∆ one has { } ∈ V j=1 ∈ Pk 2 2 Dg(φ,φ) Dg(φ, φ) = j=1 λjaj λk φ 2 λk and so supφ∈V 2 λk. On the other ≤ k kL ≤ ∆ kφkg ≤ 2 Dg(φ,φ) hand, Dg(ϕk, ϕk) = λk ϕk 2 and so supφ∈V 2 λk. It follows that k kL ∆ kφkg ≥ Dg(φ, φ) sup 2 = λk. φ∈V φ ∆ k kg Remark 61. The statement of the Max-min (resp. Min-max) Theorem holds writing an “=” sign istead of (resp. ). The proof is not too hard but we skip it. ≥ ≤ 7.3 Domain monotonicity Theorem 62 (Domain monotonicity for Dirichlet data). Let M be a compact Rie- mannian manifold with piece-wise smooth (or empty) boundary, with a given eigenvalue problem on ∂M. Let Ω1,..., Ω` M be pairwise disjoint open sets whose boundaries ⊂ are piece-wise smooth and such that any intersection with ∂M is transversal. For each i = 1, . . . , ` impose Dirichlet boundary conditions on Ωi except on ∂M Ωi where the ∩ initial data remains unchanged. Write µ1 µ2 ... for the eigenvalues of all the Ωi’s. Then, ≤ ≤ λk µk. ≤ 7.3 Domain monotonicity 89 Ω3 M Ω1 Ω2 Proof. Let ψi denote the eigenfunction of µi on the corresponding Ωψi with Dirichlet boundary conditions on ∂Ωψ . Set ψi = 0 on M Ωψ . Then ψi H1(M) for all i. Fix i \ i ∈ k N. We may make ψ1, . . . , ψk orthonormal. ∈ Pk ⊥ We claim that there exists φ = aiψi H1(M) (span ϕ1, . . . , ϕk−1 ) . The ar- i=1 ∈ ∩ { } gument is the same as in Theorem 59, it amounts to solve k 1 equations having k − unknowns. By Theorem 58 k 2 X λk φ Dg(φ, φ) = aiajDg(ψi, ψj), k kg ≤ i,j=1 and Z Dg(ψi, ψj) = ∆˜ gψi ψj ωg M · Z = ∆˜ gψi ψj ωg · Ωψj Z = ψi ∆˜ gψj ωg + 0 · Ωψj Z = µj ψi ψj ωg · Ωψj = µjδij. 2 Pk Pk 2 2 Therefore λk φ aiajDg(ψi, ψj) = a µi µk φ . k kg ≤ i,j=1 i=1 i ≤ k kg Theorem 63 (Domain monotonicity for Neumann data). Let M be a compact Rie- mannian manifold with piece-wise smooth (or empty) boundary, with a given eigenvalue problem on ∂M. Let Ω1,..., Ω` M be pairwise disjoint open sets whose boundaries ⊂ are piece-wise smooth and such that any intersection with ∂M is transversal. Assume further that ` M = Ωi. ∪i=1 For each i = 1, . . . , ` impose Neumann boundary conditions on Ωi except on ∂M Ωi ∩ where the initial data remains unchanged. Write ν1 ν2 ... for the eigenvalues of ≤ ≤ all the Ωi’s. Then, νk µk. ≤ Proof. Let ψi be the eigenfunction corresponding to νi on the appropriate Ωψi , and set ψi = 0 on M Ωψi . As before, we may assume the ψi are orthonormal and we \ Pk have ψi H1(M). As we have shown before, there exists φ = i=1 aiϕi so that ∈ ⊥ φ H1(M) (span ψ1, . . . , ψk−1 ) . Note that in particular φ H(Ωj) for all j. ∈ ∩ { } ∈ 90 Eigenvalues Ωj Write Dg : H1(Ωj) H1(Ωj) R for the Dirichlet form on Ωj. Note that by Theorem × → 58 Z Ωj 2 Dg (φ, φ) νk φ ωg. ≥ Ωj | | It then follows that ` ` Z X Ωj X 2 2 Dg(φ, φ) = Dg (φ, φ) νk φ ωg = νk φ . ≥ | | k kg j=1 j=1 Ωj On the other hand, k k X X 2 2 Dg(φ, φ) = aiajDg(ψi, ψj) = a λi λk φ . i ≤ k kg i,j=1 i=1 7.4 Simple lower bound for λk m 2 The operator (∆g +I) : H2m(M) L (M) is elliptic for all m N and so by G¨arding’s → ∈ inequality there exists C > 0 making m m φ H C (∆g + I) φ 2 . k k 2m ≤ k kL Theorem 64. Let n =dim M. For any integer m > n/4 there exist constants Cm > 0 and km > 0 such that 1 λk Cm k 2m ≥ for all k km. ≥ m m Proof. We start noticing that if ∆gϕj = λjϕj then (∆g + I) ϕj = (λj + 1) ϕj. Let M m m λ := ker (∆g + I) (λj + 1) . H m − (λj +1) ≤λ P For φ = ajϕj λ we have j ∈ H m 2 X m 2 X 2m 2 2 2 (∆g + I) φ 2 = (λj + 1) ajϕj 2 = (λj + 1) ajϕj 2 λ φ 2 k kL k kL k kL ≤ k kL j j and so m (∆g + I) φ 2 λ φ 2 . k kL ≤ k kL 0 By the Sobolev embedding, if 2m > n/2 then H2m(M) C (M). In particular, there ⊂ exists C1 > 0 for which φ ∞ C1 φ H for all φ H2m(M). k k ≤ k k 2m ∈ Putting all together, if φ λ, then there is C2 > 0 ∈ H m φ ∞ C1 φ H C2 (∆g + I) φ 2 C2λ φ 2 . k k ≤ k k 2m ≤ k kL ≤ k kL 7.5 First eigenvalue: Faber Krahn inequality 91 Fix x M and ` dim λ. Then, for any real numbers a1, . . . , a`, ∈ ≤ H ` ` ` ` 1/2 1/2 X X X 2 X 2 aj ϕj(x) C2λ aj ϕj = C2λ aj ϕj L2 = C2λ aj . ≤ L2 k k j=1 j=1 j=1 j=1 Now pick aj = ϕj(x). We then get ` ` 1/2 X 2 X 2 ϕj(x) C2λ ϕj(x) ≤ j=1 j=1 and so for all x M ∈ ` X 2 2 2 ϕj(x) C λ . ≤ 2 j=1 Integrating over M 2 2 ` C α volg(M) ≤ 2 2 2 m and therefore in particular dim λ C λ volg(M) for any λ. Picking λ := (λk + 1) H ≤ 2 we get k = dim λ and so H k m √k (λk + 1) p . ≥ C2 volg(M) 7.5 First eigenvalue: Faber Krahn inequality Given two regions with the same volume, the one with the largest boundary should be the one that looses heat the fastest. Since a solution for the heat equation is dominates −λ t by the term e 1 ϕ1, we should have that the first eigenvalue corresponding to the region with larger boundary should be greater. The results in this section are presented for n subset of R but can be rewritten with almost no modification for a compact Riemannian manifolds (M, g). n Theorem 65 (Faber Krahn inequality). Let Ω R be a bounded domain and let n ⊂ B R denote a ball satisfying vol(Ω) = vol(B). Then, ⊂ λ1(Ω) λ1(B) ≥ Where λ1(Ω) and λ1(B) are the first eigenvalues for the Dirichlet eigenvalues on Ω and B respectively. To prove this Theorem we will use in several ocassions the co-area formula: Z Z max φ Z ψ ψ dx = dτ ds. −1 φ {φ>t} t ϕ (s) |∇ | 92 Eigenvalues Proof. Let φ V and set ∈ n Ωt := x R : φ(x) > t . { ∈ } We now define a symmetrization φ∗ : B [0, + ) of φ. Let Bt be a ball centered at → ∞ the origin satisfying vol(Bt) = vol(Ωt). The symmetrization φ∗ is defined as the radially symmetric function such that 2 x R : φ∗(x) > t = Bt. { ∈ } By the co-area formula Z max φ Z 1 Z max φ∗ Z 1 dτ ds = vol(Ωt) = vol(Bt) = dτ ds. −1 φ −1 φ∗ t φ (s) |∇ | t φ∗ (s) |∇ | Differentiation with respect to t we get Z 1 Z 1 dτ = dτ for all t. (7.2) −1 φ −1 φ∗ φ (t) |∇ | φ∗ (t) |∇ | Then, Z Z max φ Z φ2 φ2 dx = dτ ds −1 φ Ω 0 φ (s) |∇ | Z max φ Z 1 = s2 dτ ds −1 φ 0 φ (s) |∇ | Z max φ Z 1 = s2 dτ ds −1 0 φ∗ (s) φ∗ Z |∇ | 2 = φ∗ dx B where the last equality follows from the fact that max φ = max φ∗. For t [0, max φ] define the functions ∈ Z Z 2 2 G(t) = φ dx and G∗(t) = φ∗ dx. Dt |∇ | Bt |∇ | By the co-area formula Z max φ Z G(t) = φ dτ ds t φ−1(s) |∇ | and so Z G0(t) = φ dτ. − φ−1(t) |∇ | Analogously, Z 0 G∗(t) = φ∗ dτ. −1 − φ∗ (t) |∇ | 7.6 Continuity of eigenvalues 93 By Cauchy-Schwartz inequality !2 ! ! Z Z Z 1 (vol(φ−1(t)))2 = 1 dτ φ dτ dτ . −1 ≤ −1 |∇ | −1 φ φ (t) φ (t) φ (t) |∇ | −1 On the other hand, since φ∗ is constant on φ (t), |∇ | ∗ Z !2 Z ! Z ! −1 2 1 (vol(φ∗ (t))) = 1 dτ = φ∗ dτ dτ . −1 −1 |∇ | −1 φ∗ φ∗ (t) φ∗ (t) φ∗ (t) |∇ | The isoperimetric inequality says that vol(φ−1(t)) vol(φ−1(t)). ≥ ∗ It follows from (7.2) that Z Z 0 0 G (t) = φ dτ φ∗ dτ = G∗(t). −1 −1 − φ (t) |∇ | ≤ − φ∗ (t) |∇ | Integrate with respect to t (using that G(max φ) = 0 = G∗(max φ)) and apply the co-area formula to get Z Z 2 2 φ dx = G(0) G∗(0) = φ∗ dx. Ω |∇ | ≥ B |∇ | It follows that R 2 R 2 Ω φ dx B φ∗ dx λ1(Ω) = R|∇ 2| R|∇ 2 | = λ1(B). Ω φ dx ≤ B φ∗ dx 7.6 Continuity of eigenvalues The notes in this section where written by Dmitri Gekhtman. Theorem 66 (Continuity in the C0-topology of metrics). Let M be a compact manifold and let g and g˜ be two Riemannian metrics on M that are close in the sense that there exists ε > 0 small making (1 ε)˜g g (1 + ε)˜g. − ≤ ≤ Then, λ (g) 1 (n + 1)ε + O(ε2) k 1 + (n + 1)ε + O(ε2). − ≤ λk(˜g) ≤ Proof. At the level of the volume measures we have that n n (1 ε) 2 ωg˜ ωg (1 + ε) 2 ωg˜ − ≤ ≤ n 2 2 n 2 and so (1 ε) 2 ϕ ϕ (1 + ε) 2 ϕ . − k kg˜ ≤ k kg ≤ k kg˜ 94 Eigenvalues ∞ 2 R Pn ij ∂ϕ ∂ϕ Note that for any ϕ C (M) we have gϕ = g ωg and therefore, ∈ k∇ kg M i,j=1 ∂xi ∂xj since (1 ε)˜g−1 g−1 (1 + ε)˜g−1, we get − ≤ ≤ n +1 2 2 n +1 2 (1 ε) 2 g˜ϕ gϕ (1 + ε) 2 g˜ϕ . − k∇ kg˜ ≤ k∇ kg ≤ k∇ kg˜ It then follows that n +1 2 2 n +1 2 (1 ε) 2 g˜ϕ gϕ (1 + ε) 2 g˜ϕ k∇ kg˜ g k∇ kg˜ − n 2 k∇ 2k n 2 . (1 + ε) 2 ϕ ≤ ϕ g ≤ (1 ε) 2 ϕ k kg˜ k k − k kg˜ We aim to show that the kth eigenvalue of the Laplacian on Riemannian manifold depends continuously on the metric. For this to make sense, we must first establish a topology on the space of metrics . Let M be a compact, connected, smooth manifold M of dimension and let 2(M) be the set of type (2, 0) tensor fields on M. Fix a finite T cover Uσ σ∈I of M by open neighborhoods, each satisfying Uσ Vσ for some open { } 2 ⊂ coordinate neighborhood Vσ. For any h (M), let hij denote the components of h ∈ T with respect to the coordinates on Vσ. For every nonnegative integer k and σ I we ∈ define X X α h k,σ = ∂ hij . k k | | |α|≤k ij We define X h k = h k,σ. k k k k σ∈I Finally, we define a norm on 2(M) by T ∞ X −k −1 h = 2 h k(1 + h k) k k k k k k k=0 0 2 2 and define d : (M) (M) R to be the associated distance T × T → 0 d (h1, h2) = h1 h2 . k − k Next, we define a distance d00 on by M 00 00 d (g1, g2) = sup dx(g1, g2), x∈M where 00 −δ δ d (g1, g2) = inf δ > 0 e (g2)x < (g1)x < e (g2)x . x { | } If A, B are inner products A < B means that B A is positive definite. − Now, we define d = d0 + d00. The distance d on is complete and defines the C∞ M topology on . M Theorem 67. Let λk(g) be the kth eigenvalue of the Laplacian associated to g. λk is a continuous function on with respect to the C∞ topology. More precisely, d(g, g0) < δ implies M 0 0 exp( (n + 1)δ)λk(g ) λk(g) exp((n + 1)δ)λk(g ). − ≤ ≤ 7.6 Continuity of eigenvalues 95 Proof. Suppose d(g, g0) < δ. Then d00(g, g0) < δ, which implies that e−σg0 < g < eδg0. Let U, (xi) be a coordinate patch on M. Then { } −σ 0 δ 0 e (gij) < (gij) < e (gij), where (gij) is the positive definite symmetric matrix defined by the components of g. If A, B are positive definite symmetric matrices satisfying A < B, then we have the estimates B < A for the determinants and B−1 < A−1 for the inverses. Hence, we | | | | have q n q 0 n q 0 exp( σ) g < gij < exp( δ) g − 2 | ij| | | 2 | ij| and −σ 0ij δ 0ij e (g ) < (gij) < e (g ). Now, suppose f is a smooth function compactly supported in U. Then we have Z q Z 2 2 n 2q 0 n 2 f g = f gij dx exp( δ) f gij dx = exp( δ) f g0 , k k U | | ≤ 2 U | | 2 k k and similarly f 2 exp( n δ) f 0 . k kg ≥ − 2 k kg If ω is a one-form compact supported in U, we have Z q Z 2 ij n ij q 0 n 2 ω g = g ωiωj gij dx exp(( + 1)δ) g ωiωj gij dx = exp(( + 1)δ) ω g0 , k k U | | ≤ 2 U | | 2 k k 2 n 2 and similarly ω exp( ( + 1)δ) ω 0 . k kg ≥ − 2 k kg Hence, we have the inequalities n 2 2 n 2 exp( ( + 1)δ) df 0 df exp(( + 1)δ) df 0 − 2 k kg ≤ k kg ≤ 2 k kg and n 2 2 n 2 exp( δ) f 0 f exp( δ) f 0 − 2 k kg ≤ k kg ≤ 2 k kg for all smooth functions f compactly supported in a coordinate neighborhood. Using a partition of unity, we find that the inequalities hold for all f in C∞(M). Combining the inequalities, we obtain 2 2 2 df 0 df df 0 k kg g k kg exp( (n + 1)δ) 2 k k2 exp((n + 1)δ) 2 . − f 0 ≤ f ≤ f 0 k kg k kg k kg Recalling 2 df g λk(g) = inf sup k k2 , V k k f f∈V \{0} k kg the last inequality yields 0 0 exp( (n + 1)δ)λk(g ) λk(g) exp((n + 1)δ)λk(g ). − ≤ ≤ This implies 0 λk(g ) λk(g) (exp((n + 1)δ) 1)λk(g), − ≤ − which proves continuity of λk. 96 Eigenvalues In the proof of the last statement, we established the inequality λk(g) exp( (n + 1)δ) 0 exp((n + 1)δ), − ≤ λk(g ) ≤ 0 λk(g) which shows that if g, g are close, 0 is close to 1 uniformly in k. As a consequence, λk(g ) we have the following Corollary 68. The multiplicity of the kth eigenvalue is an upper-semicontinuous func- 0 tion of the metric. That is, for any g , k N, there is a δ so that d(g, g ) < δ 0 0 ∈ M ∈ implies # j λj(g ) = λk(g ) # j λj(g) = λk(g) . { | } ≤ { | } 7.7 Multiplicity of eigenvalues If (M, g) is a compact boundary-less Riemannian manifold of dimension n 3, then ≥ Colin de Verdiere proved that every finite sequence 0 = λ1 < λ2 λ3 λk is ≤ ≤ · · · ≤ the sequence of the first k eigenvalues counted with multiplicity of the Laplacian. In particular, there are no restrictions on the multiplicities of eigenvalues on manifolds of dimension n 3. In contrast, this picture is very different on surfaces. Indeed, the ≥ following holds: 2 on (S , g) one has mj 2j + 1 • ≤ 2 on (P (R), g) one has mj 2j + 3 • ≤ 2 on (T , g) one has mj 2j + 4 • ≤ where g in the above examples denotes a general Riemannian metric. Observe that if one chooses the standard metrics on these surfaces then one obtains the equalities on the multiplicities. 7.8 High energy eigenvalue asymptotics Let (M, g) be a compact boundary-less Riemannian manifold. Write 0 = λ1 < λ2 λ3 ... ≤ ≤ for all the Laplace eigenvalues repeated according to their multiplicity. We begin this section by introducing the Zeta function Zg : (0, + ) R ∞ → ∞ X −λj t Zg(t) = e . j=1 Since the series is uniformly convergent on intervals of the form [t0, + ) for all t0 > 0 we ∞ know that Zg is continuous. We also have that it is decreasing in t, that limt→0+ Zg(t) = + , and limt→+∞ Zg(t) = 0. ∞ Proposition 69. 1 + Zg(t) (volg(M) + O(t)) as t 0 . ∼ (4πt)n/2 → 7.8 High energy eigenvalue asymptotics 97 Proof. ∞ X −λj t Zg(t) = e j=0 Z = p(x, x, t) ωg(x) M k Z 1 X j k+1 = n/2 t uj(x, x) ωg(x) + O(t ) (4πt) j=0 M 1 = (volg(M) + O(t)) (4πt)n/2 Let us now write 0 = ν1 < ν2 < ν3 < . . . for all the distinct eigenvalues. Then, setting mj for the multiplicity of νj we can rewrite ∞ X −νj t Zg(t) = mje . j=1 Theorem 70. The function Zg determines all the eigenvalues and their multiplicities. Proof. Note that for µ > 0 with µ = 2, 6 0 if µ < ν , ∞ 2 µt X (µ−νj )t lim e Zg(t) 1 = lim mje = + if µ < ν2 t→∞ − t→∞ j=2 ∞ m2, if µ = ν2. It follows that ν2 is the unique strictly positive real number µ such that the limit µt limt→∞ e Zg(t) 1 is a natural number. By induction, νk is the unique strictly − positive real number µ such that the limit k−1 µt X −νj t mk := lim e Zg(t) 1 mje t→∞ − − j=2 is a natural number. Theorem 71 (Karamata). Suppose that µ is a positive measure on [0, ) and that ∞ α (0, ). Then ∈ ∞ Z ∞ e−tx dµ(x) at−α t 0 0 ∼ → implies Z λ a dµ(x) λα x . 0 ∼ Γ(α + 1) → ∞ 98 Eigenvalues + α −1 + Proof. Define the measures on R by setting µt(A) := t µ(t A) for A R . Observe ⊂ that if χA is the indicator function for any set A, then by definition Z Z α χA(λ)dµt(λ) = t χA(tλ)dµ(λ). 2 + It follows that for any f L (R ) ∈ Z Z α f(λ)dµt(λ) = t f(tλ)dµ(λ), and so in particular, Z Z −λ α −tλ lim e dµt(λ) = lim t e dµ(λ) = a. ( ) t→∞ t→∞ ∗ 1 R −λ α−1 Note that by definition of the Gamma function Γ(α+1) e αλ dλ = 1. We therefore define the measure dν(λ) := αλα−1dλ and get Z Z −λ a −λ lim e dµt(λ) = e dν(λ). t→∞ Γ(α + 1) + + −λs Consider the space := span gs : R R : gs(λ) = e , s (0, + ) . By B { → ∈ ∞ } performing a change of variables one checks Z a Z lim h(λ)dµt(λ) = h(λ)dν(λ) t→∞ Γ(α + 1) for all h . By the locally compact spaces version of the Stone-Weirstrass Theorem ∈ B ∞ + ∞ + one has that is dense in C0 (R ) = f C (R ): f(λ) vanishes as λ . Let ∞ + B λ ∞ + { ∈ ∞→ ∞}+ f Cc (R ). Since f(λ)e C0 (R ), then there exists a sequence hj C0 (R ) with ∈ λ ∈ ∈ limj→∞ hj = e f. Note that for each j we get Z Z −λ a −λ lim hj(λ)e dµt(λ) = hj(λ)e dν(λ). t→∞ Γ(α + 1) In order to interchange limj→∞ and limt→0 we use that according to ( ) the measures −λ ∗ e dµt are uniformly bounded. We proved Z a Z lim f(λ)dµt(λ) = f(λ)dν(λ). t→∞ Γ(α + 1) In particular this equality holds for f = χ[0,1] and it is easy to check that acceding to our definitions of the measures µt and ν Z a Z lim χ[0,1](λ)dµt(λ) = χ[0,1](λ)dν(λ). t→∞ Γ(α + 1) is equivalent to the conclusion we desired to prove. n Let ωn be the volume of the unite ball in R , 2πn/2 ωn := . nΓ(n/2) Our aim is to prove the following Theorem: 7.9 Isospectral manifolds 99 Theorem 72 (Weyl’s asymptotic formula). Let M be a compact Riemannian manifold with eigenvalues 0 = λ0 < λ1 ... , each distinct eigenvalue repeated according to its multiplicity. ≤ Then for N(λ) := # j : λj λ , we have { ≤ } ωn n/2 N(λ) volg(M)λ , λ . ∼ (2π)n → ∞ In particular, √2π 2/n λj 2/n j , j . ∼ (ωnvolg(M)) → ∞ Proof. P For the measure µ = δλj , Proposition 69 asserts that Z ∞ −tλ 1 −n/2 e dµ(λ) n volg(M)t . 0 ∼ (4π) 2 Using Karamata’s theorem on µ and α = n/2 we obtain Z λ volg(M) n/2 ωnvolg(M) n/2 N(λ) = dµ(λ) n/2 λ = n λ . 0 ∼ (4π) Γ(n/2 + 1) (2π) 7.9 Isospectral manifolds In this section we prove that if (M, gM ) and (N, gN ) are compact Riemannian manifolds which share the same eigenvalues then they must have the same dimension, same volume and same total curvature. Theorem 73. If (M, gM ) and (N, gN ) are isospectral compact Riemannian manifolds, then Z Z dim M = dim N, volgM (M) = volgN (N) and RgM ωgM = RgN ωgN . M N Proof. Let λ0 λ1 ... be the eigenvalues of both ∆g and ∆g . Then ≤ ≤ M N ∞ k Z X −λj t 1 X j gM k+1 e = dim M/2 t uj (x, x) ωg(x) + O(t ) j=0 (4πt) j=0 M and ∞ k Z X −λj t 1 X j gN k+1 e = dim N/2 t uj (x, x) ωg(x) + O(t ) . j=0 (4πt) j=0 M 100 Eigenvalues It follows immediately that dim M = dim N := n. Next, note that Z Z 1 gM gN n/2 u0 (x, x) ωg(x) u0 (x, x) ωg(x) = (4πt) M − N k Z Z 1 X j gM gN k+1 = t u (x, x) ωg(x) u (x, x) ωg(x) + O(t ) n/2 j − j (4πt) j=1 M N yields Z Z gM gN u0 (x, x) ωg(x) = u0 (x, x) ωg(x) M N gM gN and since u0 (x, x) = u0 (x, x) = 1, we have volgM (M) = volgN (N). Repeating the same argument it follows that if (M, gM ) and (N, gN ) are isospectral compact Rieman- nian manifolds, then for all j Z Z gM gN uj (x, x) ωg(x) = uj (x, x) ωg(x). M N 1 In particular, since u1(x, x) = 6 Rg(x) we have that (M, gM ) and (N, gN ) have the same total curvature. We next prove that in the case of compact surfaces isospectrality implies a strong result: Corollary 74. If (M, gM ) and (N, gN ) are isospectral compact Riemannian surfaces, then M and N are diffeomorphic. Proof. By Gauss-Bonnet Theorem Z RgM ωgM = 8π(1 γM ) M − where γM is the genus of M. The same result holds for N. We then use that Z Z RgM ωg = RgN ωg M N which yields γM = γN . The result follows from the fact that two orientable surfaces with the same genus are diffeomorphic. Theorem 75. Suppose (M, gM ) and (N, gN ) are isospectral compact Riemannian man- ifolds of dimension n = 2, 3, 4, 5. If (M, gM ) has constant sectional curvature, then so does (N, gN ). Remark 76. In dimension n = 6 the result also holds provided we ask the sectional curvatures of (M, gM ) to be strictly positive. 2 2 Theorem 77. If (M, g) is isospectral to (S , gS2 ), then (M, g) is isometric to (S , gS2 ). CHAPTER 8 Eigenfunctions Write (x, ξ) for the coordinates of a particle in phase space. That is, x denotes position ∗ and ξ is the momentum. Let H : T M R be the Hamiltonian → 1 H(x, ξ) = ξ 2 + V (x). 2| |g As discussed in the Introduction, the classical Hamiltonian equations ( ∂xj = ∂H ∂t ∂ξj ∂ξ j = ∂H , ∂t − ∂xj describe the motion of a particle with kinetic energy 1 ξ 2 and potential energy V (x). 2 | | The idea of Schr¨odingerwas to model the behavior of the electron by a wave-function ϕj that solves the problem 2 h ∆g + V ϕj = Ej(h)ϕj. − 2 Here h is Planck’s constant, a very small number h 6.6 10−34 m2kg/s. If we choose ∼ × a system free of potential energy, V = 0, then the limit h 0 is equivalent to the high → frequency limit λ . When working with normalized eigenfunctions, ϕj 2 = 1, one → ∞ k k has that for any A M ⊂ Z 2 ϕj(x) ωg(x) = P particle of energy Ej(h)/h belongs to A . A | | The time evolution of a particle in the initial state u0 is given by 2 −i t (− h ∆ +V ) u(x, t) = e h 2 g u(x). Note that for all t 2 t h2 tEj (h) −i (− ∆g+V ) −i 2 2 e h 2 ϕj(x) dx = e h ϕj(x) dx = ϕj(x) dx | | | | 102 Eigenfunctions and so eigenstates are stationary states. We start studying the behavior of solutions ϕλ to the equation ∆gϕλ = λϕλ on balls B(x0, r) for r small. They are not necesarilly solutions on the whole manifold. Some- times local results on smalls balls can be extended to solutions on all of M by covering arguments. 8.1 Local properties of Eigenfunctions It can be shown that on small length scales comparable to wavelegth scale 1/√λ eigen- functions behave like harmonic functions. Fix an atlas over M. Then, in local coordi- nates at x0 M the Laplace equation is given by ∈ n 1 X ijp √ ∂x g det g∂x ϕλ (x) = λϕλ(x) x B (x0). −p i j ∈ / λ det g(x) i,j=1 We rescale this problem to the unit ball. That is, we set r = /√λ and given any function u on M we write ur for the rescaled function ur(x) = u(rx). Then, the Laplace equation becomes n X ijp 2p ∂x g det gr∂x ϕλ,r (x) = gr(x)ϕλ,r(x) x B1(x0). − i r j ∈ i,j=1 Consider the operator n X ijp 2 L = ∂x g det gr∂x √gr. − i r j − i,j=1 Then the Laplace equation becomes Lϕλ,r = 0 on B1(0). For ε > 0 small enough, the operator L is close to be the Euclidean Laplacian and ϕλ,r is close to being a harmonic function. This property is extensively used in the works of H. Donelly, C. Fefferman and N. Nadirashvilli. Eigenfunctions ϕλ in small length scales are an analogue to polynomials of degree √λ. Actually, polynomials and eigenfunctions have many properties in common: Order of vanishing. • Local growth. • Local structure of nodal sets. • The frequency function of a given function u C∞(M) measures the local growth rate ∈ n of u. Let u be any harmonic function on B1(0) R . For a B1(0) and 0 < r 1 a , ⊂ ∈ ≤ −| | define the frequency of u at a in the ball Br(a) by Du(a, r) Nu(a, r) = r , Hu(a, r) 8.1 Local properties of Eigenfunctions 103 where Z Z 2 2 Du(a, r) = u dx, Hu(a, r) = u dσ. Br(a) |∇ | ∂Br(a) It can be shown that Nu(a, r) is a monotone non-dedreasing function of r (0, 1 a ) for ∈ −| | any a B1(0). Supose u is a harmonic homogeneous polynomial of degree k. Then we ∈ k may write u on polar coordinates u(r, ω) = ckr φk(ω) where φk is a spherical harmonic of degree k on Sn−1. Then, 2 2k k ck r Nu(0, r) = 2 2k = k = degree(u). ck r d n−1 Since by integration by parts one can show that dr Hu(a, r) = r Hu(a, r) + 2Du(a, r), we get d Hu(a, r) 2Nu(a, r) log = . dr rn−1 r We therefore obtain that for all 0 < R < 1 (1 a ), 2 − | | Z 2R Hu(a, 2R) Hu(a, R) Hu(a, R) N(a,1−|a|) n−1 = n−1 exp 2Nu(0, r)/r dr n−1 4 (2R) R R ≤ R In particular, Hu(a, λR) Hu(a, R) = λ2Nu(0,1) 1 λ 2 ( ) (λR)n−1 Rn−1 ≤ ≤ ∗ Integrating with respect to R this gives 1 Z 1 Z u2dx λ2Nu(0,1) u2dx 0 R 1/2, 1 λ 2. vol(BλR(0)) BλR(0) ≤ vol(BR(0)) BR(0) ≤ ≤ ≤ ≤ Using this one can prove that the order of vanishing νu(a) of u at a B (0) is bounded ∈ 1/4 νu(a) CNu(0, 1) + c(n). ≤ Using that ϕλ,r(x) = ϕλ(rx) is almost harmonic the previous order of vanishing estimate can be extended to manifolds. But in order to do that one has to extend the definition of the frequency function. It turns out that on a compact manifold (M, g) the natural extension of the frequency function of a harmonic map u C∞(M) is ∈ Du(a, r) Nu(a, r) = r , Hu(a, r) where Z Z 2 2 Du(a, r) = µ u dx, Hu(a, r) = µu dσ Br(a) |∇ | ∂Br(a) with gij(x)xixj µ(x) = . x 2 | | One then gets the followin result on the order of vanishing of ϕλ: 104 Eigenfunctions Theorem 78 (Order of vanishing). Let (M, g) be a compact manifold of dimension n. Then there exists C > 0 such that νϕ (a) C√λ for all a M. λ ≤ ∈ Using that ϕλ,r(x) = ϕλ(rx) is almost harmonic the previous doubling estimate can be extended to the following result. Theorem 79 (Doubling estimate). Let ϕλ be a global eigenfunction for ∆g on (M, g) compact. Then there exists C > 0 and r0 > 0 such that for all 0 < r < r0 Z √ Z 1 2 C λ 1 2 ϕλ ωg e ϕλ ωg. vol(B2r(a)) B2r(a) | | ≤ vol(Br(a)) Br(a) | | In addition, for 0 < r0 < r, √ r C λ max ϕλ(x) 0 max ϕλ(x) . x∈Br(p) | | ≤ r x∈Br0 (p) | | 8.1.1 Gradient estimates n Harmonic functions on R satisfy many other nice properties. For instance, they satisfy n Bernstein type estimates. Let u be harmonic on R . Then C sup u(x) sup u(x) x∈Br(a) |∇ | ≤ r x∈B2r(a) | | and Z 2 C 2 sup u(x) n+2 u(x) dx. x∈Br(a) |∇ | ≤ r Br(a) | | The translation of these results to the setting of eigenfunctions on compact manifolds is the following. Theorem 80 (Bernstein inequalities). Let (M, g) be a compact manifold. Then, 1 Cλ 4 −1/8 sup ϕλ(x) sup ϕλ(x) , r C2λ x∈Br(a) |∇ | ≤ r x∈B2r(a) | | ≤ and Z Z 2 Cλ 2 ϕλ(x) ωg(x) 2 ϕλ(x) ωg(x). Br(a) |∇ | ≤ r Br(a) | | 8.2 Global properties of eigenfunctions 105 8.1.2 Positive mass on subsets of M Suppose that (M, g) is compact and that ϕλ is a global eigenfunction, ∆gϕλ = λϕλ. Then, if A M, ⊂ Z √ 2 −C λ ϕλ(x) ωg(x) c e . A | | ≥ √ ` −c λdg(x,γ) As an illustration, the highest weight spherical harmonics Y` decay at a rate e away from a stable elliptic orbit γ, where λ = `(` + 1). A semi-classical lacuna is an open subset A M for which there exist a sequence 2 ⊂ ϕλ of L -normalized eigenfunctions and constants c, C > 0 so that { jk } Z 2 −C√λj ϕ (x) ωg(x) c e k . λjk A | | ≤ Another descriptive term is exponential trough. Lacunae are also known as classically ` forbidden regions. For instance, on the sphere the sequences Ym with m/` E 2 { } → concentrates on an invariant annulus KE S , which is known as the “classically ⊂ ` −`d (x,K ) allowed region”. One can show that for this sequence Y (x) e A E , where da | m | ≤ denotes the Agmon distance. It is unknown whether semi-classical lacunas can occur on (M, g) with classically chaotic (i.e. highly ergodic) geodesic flows. In contrast, we have the following result: Theorem 81 (Quantum ergodicity). If (M, g) is a compact manifold with ergodic geodesic flow then there exists a density one subsequence of eigenfunctions ϕj k such { k } that for any A M ⊂ Z 2 volg(A) lim ϕjk (x) ωg(x) = . k→∞ A | | volg(M) #{k: jk≤m} By density one subsequence it is meant that infm m = 1. This result is due to Schnirelman (1973) finished by Colin de Verdiere (1975). Remark. On arithmetic surfaces the above result holds for the entire sequence of eigenfunctions. This is known as Quantum Unique Ergodicity. 8.2 Global properties of eigenfunctions 8.2.1 L∞-norms Weyl’s law says that ωnvolg(M) N(λ) = # eigenvalues λ λn/2 { ≤ } ∼ (2π)n and one can further prove what’s known as local Weyl’s law X 2 ωnvolg(M) n/2 ϕλ (x) = λ + R(λ, x) | j | (2π)n λj ≤λ 106 Eigenfunctions n−1 with R(λ, x) = O(λ 2 ) uniformly in x M. In particular, ∈ X 2 n−1 ϕj(x) = O(λ 2 ) | | λj =λ uniformly in x M. In particular, the following result holds: ∈ Theorem 82. Let (M, g) be a compact Riemannian manifold of dimension n. Then, if ∆gϕλ = λϕλ, n−1 ϕλ L∞ = O(λ 4 ). k k Let Vλ = ker(∆g λ) and set − X Πλ(x, y) := ϕλj (x)ϕλj (y). j: λj ≤λ We define the coherent state at x M by ∈ x Πλ(x, y) Φλ(y) := p , y M. Πλ(x, x) ∈ The coherent states are extremes for the L∞-norms. Indeed, Z sZ 2 p x ϕλ(x) = Πλ(x, y)ϕλ(y)ωg(y) Πλ(x, y) ωg(y) Πλ(x, x) = Φλ(x) . | | M ≤ M | | ≤ | | ` Remark. On the sphere, the zonal spherical harmonic Y0 of degree ` is the coherent x state Φλ where x is the north pole and λ = `(` + 1). The L∞-norm estimate is very rarely sharp. Here we show that if the upper bound is attained, then there must be a recurrent point on M. For x M consider the set x ∗ ∈ L of ξ SxM that are the initial velocities of geodesic loops that start at x. That is, ∈ ∗ x := ξ SxM : expx(T ξ) = x for some T . We write x for its measure. For L { ∈ 2 } |L | example, on (S , g 2 ) we have x = 2π for x being the south or north pole. S |L | n−1 Note that we know sup ϕ L∞ : ϕ Vλ, ϕ = 1 = O(λ 4 ). {k k ∈ k k } Theorem 83. Suppose n−1 sup ϕ L∞ : ϕ Vλ, ϕ = 1 is not o(λ 4 ). {k k ∈ k k } Then, there exists x M for which x > 0. In particular, if g is analytic, then all the ∈ |L | geodesics loops at x must return to x at the same time. The proof of this result is based on the study of R(λ, x). Indeed, if x = 0, then n−1 |L | R(λ, x) = ox(λ 2 ). It follows that there are topological restrictions for M to have a real analytic metric such that some sequence of eigenfunctions has the maximal sup-norms. Among all pos- sible surfaces, only the sphere possesses such a metric. In addition, a metric on S2 8.2 Global properties of eigenfunctions 107 with ergodic geodesic flow can never exhibit the maximal growth rate achieved by zonal 2 harmonics on (S , gS2 ). When the geodesic flow is chaotic, the random wave conjecture predicts that eiegen- functions should behave like Gaussian random functions. In particular, one should have ϕλ L∞ = O(√log λ). This is very likely to be true in most chaotic systems but not for k k all of them. Indeed, It has been shown that on some special arithmetic hyperbolic quo- tients the O(√log λ) doesn’t hold. The best result known to date if that on manifolds n−1 with no conjugate points ϕλ L∞ = O(λ 4 / log λ). k k 8.2.2 Lp-norms. In general Lp-norms are hard to compute. For general Lp-norms one has the following general result due to C. Sogge. Theorem 84. Let (M, g) be compact Riemannian manifold and let ϕλ be a normalized eigenfunction of eigenvalue λ. Then, for all 2 p ≤ ≤ ∞ δ(p) ϕλ Lp = O(λ 2 ) k k where ( n( 1 1 ) 1 if 2(n+1) p , δ(p) = 2 − p − 2 n−1 ≤ ≤ ∞ . n−1 ( 1 1 ) if 2 p 2(n+1) . 2 2 − p ≤ ≤ n−1 2(n+1) The upper bounds are saturated on the round sphere. For p > n−1 , zonal (rotation- ally invariant) spherical harmonics saturate the Lp-bounds. For 2 p 2(n+1) the ≤ ≤ n−1 bounds are saturated by highest weight spherical harmonics, i.e. Gaussian beam along a stable elliptic geodesic. The zonal has high Lp norm due to its high peaks on balls of radius 1/√λ . The balls are so small that they do not have high Lp norms for small p. The Gaussian beams are not as high but they are relatively high over an entire geodesic. 1 Sogge’s result holds for p 2. For the L we have the obvious bound ϕλ L1 ≥ p k k ≤ ϕλ 2 1. It is interesting however to observe that we can use the L bounds for k kL ≤ p 2 to get a lower bound on ϕλ 1 . ≥ k kL Theorem 85. Let (M, g) be compact Riemannian manifold and let ϕλ be a normalized eigenfunction of eigenvalue λ. Then, there exists c such that − n−1 ϕλ 1 cλ 8 . k kL ≥ 2(n+1) p Proof. Fix 2 < p . Sogge’s L bounds give that there exists Cp > 0 for which ≤ n−1 n−1 ( 1 − 1 ) ϕλ Lp Cpλ 4 2 p . k k ≤ By H¨older’sinequality, 1 1 θ θ −1 ϕλ 2 ϕλ 1 ϕλ 2 k kL ≤ k kL k kL 108 Eigenfunctions for θ = p 1 1 = p−2 . It follows that p−1 2 − p 2(p−1) 1 1 1 −1 n−1 1 1 θ −1 θ θ 4 ( 2 − p ) 1 = ϕλ 2 ϕλ 1 ϕλ 2 ϕλ 1 Cpλ . k kL ≤ k kL k kL ≤ k kL The result follows from observing that n−1 1 1 1 1 = n−1 . 4 2 − p θ − 8 Symmetry of Lp-norms. Jakobson and Nadirashvilli studied the relation between the Lp norm of the negative and positive parts of eigenfunctions. Indeed, for χ the indicator function, set + − ϕ := ϕ χ , and ϕ := ϕλ χ . λ · {ϕλ≥0} λ · {ϕλ≤0} + Theorem 86. Let (M, g) be a compact smooth manifold. Then, for any p Z there ∈ exists Cp > 0 such that for any non constant eigenfunction ϕλ of the Laplacian + 1 ϕλ Lp k −k Cp Cp ≤ ϕ Lp ≤ k λ k for all λ. 8.3 Zeros of eigenfunctions Let (M, g) be a compact Riemannian manifold. Given any function φ C∞(M) we ∈ define its nodal set φ := x M : φ(x) = 0 . N { ∈ } Each connected component of the complement of φ is called a nodal domain. N We continue to write λ1 λ2 ... for the eigenvalues of the Laplacian repeated ac- ≤ ≤ cording to multiplicity (for any initial problem). Write ϕ1, ϕ2,... for the corresponding L2-normalized eigenfunctions. Theorem 87 (Courant’s nodal domain Theorem). The number of nodal domains of ϕk is strictly smaller than k + 1. Proof. Suppose that ϕk has at least k + 1 nodal domains D1,...,Dk+1,... and define ( ϕk Dj on Dj, ψj = | . 0 else Pk There exists φ = ajψj H1(M) orthogonal to ϕ1, . . . , ϕk−1 by the same argument j=1 ∈ in Theorem 58. Then, Dg(φ, φ) λk . ≤ φ 2 k kg Also, k X Dg(φ, φ) = aiaj ∆˜ gψi, ψj g, h i ij=1 8.3 Zeros of eigenfunctions 109 and Z Z ˜ ˜ ∆gψi, ψj g = ∆gψi ψj ωg = ∆gψi ψj ωg h i M · Dj · Z = ψi ∆gϕk ωg + 0 Dj · Z = δij ϕk ∆gϕk ωg Dj · Z 2 = λkδij ϕk ωg. Dj Therefore, k Z X 2 2 2 Dg(φ, φ) = λk a ϕ ωg λk φ . j k ≤ k kg j=1 Dj We proved Dg(φ, φ) = λ φ 2 k k kg and so it follows that φ is an eigenfunction of eigenvalue λk. Since φ vanishes on an open set Dk+1 we conclude from the unique continuation principle that φ = 0, which is a contradiction. Important remarks. ϕ1 always has constant sign. • The multiplicity of λ1 is 1: Otherwise, if ϕ2 is an eigenfunction for λ1 we would • know that ϕ2 has constant sign. Since ϕ1 has constant sign as well we have a contradiction from the fact that ϕ1, ϕ2 = 0. h ig ϕ2 has precisely two nodal domains and ϕk has at least two nodal domains for all • k 2: Otherwise ϕk has constant sign but ϕ1, ϕk = 0 . ≥ h ig Theorem 88. For any (M, g) there exists a constant C > 0 so that every ball of radius bigger that C/√λ contains a zero of any eigenfunction ϕλ. Proof. Fix x M and r > 0. Suppose that ϕλ has no zeros in Br(x). Then, there must ∈ exist a nodal domain Dλ of ϕλ such that Br(x) Dλ. Consider now the eigenvalue ⊂ problem on D λ ( ∆gφ = µφ in Dλ ( ) . ∗ φ(x) = 0 x ∂Dλ. ∈ It follows that ϕλ is an eigenfunction for ( ). In addition, since ϕλ doesn’t change sign ∗ in Dλ we must have that it is the first eigenfunction for ( ). Let us write λ1(Dλ) for ∗ the corresponding eigenvalue. By Domain monotonicity, since Br(x) Dλ, we get ⊂ λ = λ1(Dλ) λ1(Br(x)). ≤ 110 Eigenfunctions 2 2 To finish the proof one needs to show that λ1(Br(x)) C /r for some constant C > 0. ≤ Consider the Euclidean metric ge(y) := g(x) for all y Bar(x; g) where a (0, 1) is ∈ ∈ chosen so that Bar(x; ge) Br(x; g). By Domain monotonicity we get ⊂ λ1(Br(x; g)) λ1(Bar(x; g)). ≤ In addition, by comparing Rayleigh’s quotients we get that there exists C1 > 0 making λ1(Bar(x; g)) C1λ1(Bar(x; ge)). ≤ On the other hand, by explicit computations using the Bessel functions on euclidean balls it is possible to get C2 λ1(Bar(x; ge)) . ≤ r2 C1C2 It follows that λ1(Br(x; g)) 2 . ≤ r 8.3 Zeros of eigenfunctions 111 One way of measure the local asymmetry of the nodal sets is the following result due to Mangoubi. Proposition 89. Let (M, g) be a compact manifold of dimension n. Then there exists C > 0 with volg ϕλ > 0 B2r(x) − n−1 { } ∩ Cλ 2 volg(B2r(x)) ≥ for all x M and r > 0 such that ϕλ = 0 Br(x) = 0. ∈ { } ∩ 6 Using this result Mangoubi showed the following theorem on the inner radius of the nodal domains. By inner radius of D, inrad(D) we mean the largest r such that there exists a ball of radius r that can be inscribed in D. Theorem 90. Let (M, g) be a compact manifold of dimension n. Then there exists C1,C2 > 0 such that for Dλ nodal domain of ϕλ one has C1 C2 inrad(Dλ) λα(n) ≤ ≤ √λ where α(n) = 1 (n 1) + 1 . 4 − 2n Note that on surfaces this means that the inner radius of Dλ is comparable to 1/√λ. C/λ Figure: Nodal set of a bitorus 112 Eigenfunctions Hausdorff measure. We continue to write n = dimM. Let A M be any subset and ⊂ set ∞ d X d := inf (diam(Uj)) : A jUj, diamUj < δ , Hδ ⊂ ∪ j=1 and d d δ (A) := lim δ (A). H δ→0 H Since d(A) is monotone decreasing with δ we have that d(A) is well defended. How- Hδ H ever, it can be infinite. It can be shown that for A M borel set one has n(A) is ⊂ 1 H proportional to volg(A). Also, if γ M is a curve, one has that (γ) is propositional ⊂ H to the length of γ. Similarly, 0(A) is the number of points in A. H Uhlembeck proved that 0 is a regular value of the eigenfunctions for a generic set of metrics. In particular, generically, the nodal set ϕλ = 0 is a smooth hypersurface. { } In addition, it was proved by Baer that for x0 M there exists local coordinates ∈ (x1, . . . , xn) at x0 = (0,..., 0) such that an eigenfunction ϕ can be rewritten as k−1 k X j ϕ(x) = v(x) x1 + x1 uj(x2, . . . , xn) j=0 where ϕ vanishes to order k at x0, uj vanishes to order k j at (x2, . . . , xn) = (0,..., 0), − and v(x) = 0 close to p. It follows that nodal sets are rectifiable and therefore n−1 6 ( ϕ ) < . H N λ ∞ n−1 1 In 1978 J.Br¨uningproved that on surfaces there exists C > 0 for which ( ϕ ) λ 2 . H N λ ≥ Later, in 1982, S. T. Yau conjectured that on any compact n-dimensional manifold there exist constants C, c > 0 for which 1 n−1 1 cλ 2 ( ϕ ) Cλ 2 . ≤ H N λ ≤ Yau’s conjecture was proved for n-dimensional manifolds with real analytic metrics by H. Donelly and C. Fefferman in 1988. For manifolds with smooth metrics the conjecture remains open. The best known upper bound is due to R. Hardt and L. Simon (1989): Theorem 91. Let (M, g) be a compact n-dimensional Riemannian manifold. Then, there exists C > 0 for which √ n−1 C λ ( ϕ ) λ . H N λ ≤ The best known lower bound to date is due to T. Colding and P. Minicozzi (2010): Theorem 92. Let (M, g) be a compact n-dimensional Riemannian manifold. Then, there exists c > 0 for which 3−n n−1 cλ 4 ( ϕ ). ≤ H N λ The same result was obtained in a very neat proof by H. Hezari and C.Sogge. This proof is heavily based on a result by C.Sogge and S. Zelditch which we prove next. 8.4 Random wave conjecture 113 Proposition 93. Let (M, g) be a compact n-dimensional Riemannian manifold. For any f C2(M) ∈ Z Z (∆gf λf) ϕλ ωg = 2 f gϕλ σg. − | | − |∇ | M Nϕλ 2 2 1 s/2 Choosing f = (1 + λϕ + gϕλ ) 2 and using the Sobolev bounds ϕλ H = O(λ ) λ |∇ |g k k s H. Hezari and C.Sogge proved as a corollary of Proposition 93 that there exists c > 0 making 2 n−1 c √λ ϕλ 1 ( ϕ ). (8.1) k kL ≤ H N λ Inequality (8.1) cannot be improved on general manifolds since it is saturated by zonal n−1 harmonics for which one can check that ( Y ` ) √λ. H N 0 ∼ 1 1−n Combining (8.1) with the L - lower bound cλ 8 ϕλ 1 that we proved in Theorem ≤ k kL 85 we get the proof of Theorem 92.We proceed to prove Proposition 93. 1 N+(λ) Proof of Proposition 93. Given λ write D+(λ),...,D+ (λ) for all the positive nodal 1 N−(λ) domains of ϕλ, and analogously write D−(λ),...,D− (λ) for the negative ones. We then have N+(λ) N−(λ) [ j [ j M = D (λ) D (λ) ϕ . + ∪ − ∪ N λ j=1 j=1 Suppose 0 is a regular value of ϕλ. Then, all the nodal domains have smooth boundary. In particular, Z Z (∆gf λf) ϕλ ωg = (∆gf λf) ϕλ ωg j − | | j − D+(λ) D+(λ) Z Z Z = f(∆g λ)ϕλ ωg + f ∂νϕλ σg ϕλ ∂νf σg j − j − j D+(λ) ∂D+(λ) ∂D+(λ) Z = f ϕλ σg. − j |∇ | ∂D+(λ) j In the last equality we used that ∂νϕλ = gϕλ and that D (λ) is a positive nodal −|∇ | + domain and so ν and gϕλ point in opposite directions. Similarly, ∇ Z Z (∆gf λf) ϕλ ωg = f ϕλ σg. k − | | − k |∇ | D−(λ) ∂D−(λ) j k Adding these identities over j and k and using that ϕ = j∂D (λ) = k∂D (λ) N λ ∪ + ∪ + we get the desired result. If 0 is not a regular value, a version of Green’s identities for domains with rough boundaries yields the same formula. 8.4 Random wave conjecture In 1977, M. Berry proposed that random linear combinations of planar waves of a fixed high frequency λ in dimension n N 1 X n Wλ(x) := ai cos (ki x + εi) , x R (8.2) √ · ∈ N i=1 114 Eigenfunctions serve as a model for high frequency wavefunctions in any quatum system in which the underlying classical dynamics is chaotic. In equation (8.2) the coefficients ai are independent standard Gaussian random variables, ki are uniformly distributed on the n sphere of radius λ in R , and εi are independent and uniformly distributed in (0, 2π]. R. Aurich, A. Becker, R. Schubert, and M. Taglieber show that Wλ lim sup k ∞k 3 λ7→∞ √n log λ ≤ almost surely. On compact manifolds one cannot reproduce this construction. Actually, we know that eigenvalues are generically simple so it is pointless to even think about considering ran- dom linear combinations of eigenfunctions with a fixed eigenvalue. S. Zelditch then proposed to consider random linear combinations of eigenfunctions with eigenvalues in a window (λ, λ + 1]. These combinations are known as Gaussian random waves: A Gaussian random wave of frequency λ on (M, g) is a random function φλ λ defined ∈ H by X φλ := ajϕj, λj ∈(λ,λ+1] −2 where aj N(0, k ) are independent and identically distributed. ∼ λ Remark 94. The normalizing constant kλ is chosen so that E( φλ 2) = 1. Also, the k k law of φλ is independent of the choice of a particular orthonormal basis. Aside from L∞-norms, much work has been done on the distribution of the nodal sets for random waves. For instance, S. Zelditch shows that the n 1 dimensional Hausdorff − measureof the nodal set satisfies Yau’s conjecture in average √ n−1 √ c λ E( ( φλ )) C λ ≤ H N ≤ for some constants c, C > 0 as long as (M, g) is either aperiodic or Zoll. The random wave model predicts that the behavior of deterministic sequences of L2- normalized eigenfunctions ϕλ as λj should coincide with the behavior of the j 7→ ∞ random plane waves Wλ. It has been shown by N. Burq and G. Lebeau that p φλ ∞ = O log λj k j k and φλ p = O(1). k j k